3.39 \(\int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=229 \[ -\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {9 a b^2 \sec (c+d x)}{2 d}-\frac {9 a b^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d} \]

[Out]

-3/8*a^3*arctanh(cos(d*x+c))/d-9/2*a*b^2*arctanh(cos(d*x+c))/d+3*a^2*b*arctanh(sin(d*x+c))/d+3/2*b^3*arctanh(s
in(d*x+c))/d-3*a^2*b*csc(d*x+c)/d-3/2*b^3*csc(d*x+c)/d-3/8*a^3*cot(d*x+c)*csc(d*x+c)/d-a^2*b*csc(d*x+c)^3/d-1/
4*a^3*cot(d*x+c)*csc(d*x+c)^3/d+9/2*a*b^2*sec(d*x+c)/d-3/2*a*b^2*csc(d*x+c)^2*sec(d*x+c)/d+1/2*b^3*csc(d*x+c)*
sec(d*x+c)^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3517, 3768, 3770, 2621, 302, 207, 2622, 288, 321} \[ -\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}+\frac {9 a b^2 \sec (c+d x)}{2 d}-\frac {9 a b^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(-3*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (9*a*b^2*ArcTanh[Cos[c + d*x]])/(2*d) + (3*a^2*b*ArcTanh[Sin[c + d*x]])
/d + (3*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^2*b*Csc[c + d*x])/d - (3*b^3*Csc[c + d*x])/(2*d) - (3*a^3*Cot[
c + d*x]*Csc[c + d*x])/(8*d) - (a^2*b*Csc[c + d*x]^3)/d - (a^3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d) + (9*a*b^2*S
ec[c + d*x])/(2*d) - (3*a*b^2*Csc[c + d*x]^2*Sec[c + d*x])/(2*d) + (b^3*Csc[c + d*x]*Sec[c + d*x]^2)/(2*d)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 3517

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \left (a^3 \csc ^5(c+d x)+3 a^2 b \csc ^4(c+d x) \sec (c+d x)+3 a b^2 \csc ^3(c+d x) \sec ^2(c+d x)+b^3 \csc ^2(c+d x) \sec ^3(c+d x)\right ) \, dx\\ &=a^3 \int \csc ^5(c+d x) \, dx+\left (3 a^2 b\right ) \int \csc ^4(c+d x) \sec (c+d x) \, dx+\left (3 a b^2\right ) \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx+b^3 \int \csc ^2(c+d x) \sec ^3(c+d x) \, dx\\ &=-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {1}{4} \left (3 a^3\right ) \int \csc ^3(c+d x) \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}+\frac {1}{8} \left (3 a^3\right ) \int \csc (c+d x) \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (9 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {9 a b^2 \sec (c+d x)}{2 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (9 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {9 a b^2 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {9 a b^2 \sec (c+d x)}{2 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.22, size = 1229, normalized size = 5.37 \[ -\frac {a^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3 \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 \left (a^3+4 b^2 a\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {a^2 b \cos ^3(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\left (-2 \cos \left (\frac {1}{2} (c+d x)\right ) b^3-7 a^2 \cos \left (\frac {1}{2} (c+d x)\right ) b\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \csc \left (\frac {1}{2} (c+d x)\right )}{4 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {a^2 b \cos ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3}{8 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {a^3 \cos ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3}{64 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a b^2 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 \left (a^3+4 b^2 a\right ) \cos ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3}{32 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 \left (a^3+12 b^2 a\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^3}{8 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 \left (b^3+2 a^2 b\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^3}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 \left (a^3+12 b^2 a\right ) \cos ^3(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^3}{8 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 \left (b^3+2 a^2 b\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^3}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a b^2 \cos ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\cos ^3(c+d x) \sec \left (\frac {1}{2} (c+d x)\right ) \left (-2 \sin \left (\frac {1}{2} (c+d x)\right ) b^3-7 a^2 \sin \left (\frac {1}{2} (c+d x)\right ) b\right ) (a+b \tan (c+d x))^3}{4 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 a b^2 \cos ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(3*a*b^2*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + ((-7*a^2*b*Cos[(c +
d*x)/2] - 2*b^3*Cos[(c + d*x)/2])*Cos[c + d*x]^3*Csc[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(4*d*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^3) - (3*(a^3 + 4*a*b^2)*Cos[c + d*x]^3*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^3)/(32*d*(a*
Cos[c + d*x] + b*Sin[c + d*x])^3) - (a^2*b*Cos[c + d*x]^3*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d
*x])^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a^3*Cos[c + d*x]^3*Csc[(c + d*x)/2]^4*(a + b*Tan[c + d*x]
)^3)/(64*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*(a^3 + 12*a*b^2)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2]]*(a
+ b*Tan[c + d*x])^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*(2*a^2*b + b^3)*Cos[c + d*x]^3*Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*(a^3 + 12
*a*b^2)*Cos[c + d*x]^3*Log[Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)
+ (3*(2*a^2*b + b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*
Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*(a^3 + 4*a*b^2)*Cos[c + d*x]^3*Sec[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^
3)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (a^3*Cos[c + d*x]^3*Sec[(c + d*x)/2]^4*(a + b*Tan[c + d*x])^3)
/(64*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(4*d*(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*a*b^2*Cos[c + d*x]^3*Sin[(c + d*x)/2]*(a +
b*Tan[c + d*x])^3)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (b^3*Cos[c
+ d*x]^3*(a + b*Tan[c + d*x])^3)/(4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x]
)^3) - (3*a*b^2*Cos[c + d*x]^3*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (Cos[c + d*x]^3*Sec[(c + d*x)/2]*(-7*a^2*b*Sin[(c + d*x)/2] - 2*b^3*
Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^3)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a^2*b*Cos[c + d*x]^3*Se
c[(c + d*x)/2]^2*Tan[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 427, normalized size = 1.86 \[ \frac {6 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 48 \, a b^{2} \cos \left (d x + c\right ) - 10 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, {\left ({\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left ({\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, {\left (3 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + b^{3} - 4 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{6} - 2 \, d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(6*(a^3 + 12*a*b^2)*cos(d*x + c)^5 + 48*a*b^2*cos(d*x + c) - 10*(a^3 + 12*a*b^2)*cos(d*x + c)^3 - 3*((a^3
+ 12*a*b^2)*cos(d*x + c)^6 - 2*(a^3 + 12*a*b^2)*cos(d*x + c)^4 + (a^3 + 12*a*b^2)*cos(d*x + c)^2)*log(1/2*cos
(d*x + c) + 1/2) + 3*((a^3 + 12*a*b^2)*cos(d*x + c)^6 - 2*(a^3 + 12*a*b^2)*cos(d*x + c)^4 + (a^3 + 12*a*b^2)*c
os(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) + 12*((2*a^2*b + b^3)*cos(d*x + c)^6 - 2*(2*a^2*b + b^3)*cos(d*x +
c)^4 + (2*a^2*b + b^3)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 12*((2*a^2*b + b^3)*cos(d*x + c)^6 - 2*(2*a^2*
b + b^3)*cos(d*x + c)^4 + (2*a^2*b + b^3)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 8*(3*(2*a^2*b + b^3)*cos(d*
x + c)^4 + b^3 - 4*(2*a^2*b + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6 - 2*d*cos(d*x + c)^4 + d*co
s(d*x + c)^2)

________________________________________________________________________________________

giac [A]  time = 5.33, size = 373, normalized size = 1.63 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 32 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, {\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 96 \, {\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 24 \, {\left (a^{3} + 12 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {64 \, {\left (b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {50 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/64*(a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 8*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*a*b^2*ta
n(1/2*d*x + 1/2*c)^2 - 120*a^2*b*tan(1/2*d*x + 1/2*c) - 32*b^3*tan(1/2*d*x + 1/2*c) + 96*(2*a^2*b + b^3)*log(a
bs(tan(1/2*d*x + 1/2*c) + 1)) - 96*(2*a^2*b + b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 24*(a^3 + 12*a*b^2)*lo
g(abs(tan(1/2*d*x + 1/2*c))) + 64*(b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d*x + 1/2*c)^2 + b^3*tan(1/2*d
*x + 1/2*c) + 6*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - (50*a^3*tan(1/2*d*x + 1/2*c)^4 + 600*a*b^2*tan(1/2*d*x
+ 1/2*c)^4 + 120*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 32*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*a^3*tan(1/2*d*x + 1/2*c)^2
+ 24*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.48, size = 254, normalized size = 1.11 \[ -\frac {a^{3} \cot \left (d x +c \right ) \left (\csc ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{3} \cot \left (d x +c \right ) \csc \left (d x +c \right )}{8 d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {a^{2} b}{d \sin \left (d x +c \right )^{3}}-\frac {3 a^{2} b}{d \sin \left (d x +c \right )}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {3 b^{2} a}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {9 b^{2} a}{2 d \cos \left (d x +c \right )}+\frac {9 b^{2} a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}+\frac {b^{3}}{2 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3 b^{3}}{2 d \sin \left (d x +c \right )}+\frac {3 b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x)

[Out]

-1/4*a^3*cot(d*x+c)*csc(d*x+c)^3/d-3/8*a^3*cot(d*x+c)*csc(d*x+c)/d+3/8/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-1/d*a^2
*b/sin(d*x+c)^3-3/d*a^2*b/sin(d*x+c)+3/d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))-3/2/d*b^2*a/sin(d*x+c)^2/cos(d*x+c)+9
/2/d*b^2*a/cos(d*x+c)+9/2/d*b^2*a*ln(csc(d*x+c)-cot(d*x+c))+1/2/d*b^3/sin(d*x+c)/cos(d*x+c)^2-3/2/d*b^3/sin(d*
x+c)+3/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 250, normalized size = 1.09 \[ \frac {a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, a b^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 4 \, b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8 \, a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16*(a^3*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 3*log(cos(d*x + c)
+ 1) + 3*log(cos(d*x + c) - 1)) + 12*a*b^2*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3*log(c
os(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 4*b^3*(2*(3*sin(d*x + c)^2 - 2)/(sin(d*x + c)^3 - sin(d*x + c))
- 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 8*a^2*b*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*lo
g(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d

________________________________________________________________________________________

mupad [B]  time = 4.04, size = 698, normalized size = 3.05 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3}{8}+\frac {3\,a\,b^2}{8}\right )}{d}-\frac {\mathrm {atan}\left (\frac {\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^3}{4}+9\,a\,b^2\right )+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )-6\,a^2\,b-3\,b^3\right )\,1{}\mathrm {i}-\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^3}{4}+9\,a\,b^2\right )+6\,a^2\,b+3\,b^3\right )\,1{}\mathrm {i}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (36\,a^4\,b^2+36\,a^2\,b^4+9\,b^6\right )+27\,a\,b^5+\frac {9\,a^5\,b}{2}-\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^3}{4}+9\,a\,b^2\right )+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )-6\,a^2\,b-3\,b^3\right )-\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {3\,b^3}{2}\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^3}{4}+9\,a\,b^2\right )+6\,a^2\,b+3\,b^3\right )+\frac {225\,a^3\,b^3}{4}}\right )\,\left (a^2\,b\,6{}\mathrm {i}+b^3\,3{}\mathrm {i}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a^3}{2}+6\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a^3+102\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {15\,a^3}{4}+108\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (26\,a^2\,b+8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (30\,a^2\,b-8\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (58\,a^2\,b+32\,b^3\right )+\frac {a^3}{4}+2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {15\,a^2\,b}{8}+\frac {b^3}{2}\right )}{d}+\frac {3\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+12\,b^2\right )}{8\,d}-\frac {a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^3/sin(c + d*x)^5,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^4)/(64*d) + (tan(c/2 + (d*x)/2)^2*((3*a*b^2)/8 + a^3/8))/d - (atan(((3*a^2*b + (3*b^3)
/2)*(tan(c/2 + (d*x)/2)*(9*a*b^2 + (3*a^3)/4) + 6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - 6*a^2*b - 3*b^3)*
1i - (3*a^2*b + (3*b^3)/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - tan(c/2 + (d*x)/2)*(9*a*b^2 + (3*a^3)
/4) + 6*a^2*b + 3*b^3)*1i)/(2*tan(c/2 + (d*x)/2)*(9*b^6 + 36*a^2*b^4 + 36*a^4*b^2) + 27*a*b^5 + (9*a^5*b)/2 -
(3*a^2*b + (3*b^3)/2)*(tan(c/2 + (d*x)/2)*(9*a*b^2 + (3*a^3)/4) + 6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) -
6*a^2*b - 3*b^3) - (3*a^2*b + (3*b^3)/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - tan(c/2 + (d*x)/2)*(9*
a*b^2 + (3*a^3)/4) + 6*a^2*b + 3*b^3) + (225*a^3*b^3)/4))*(a^2*b*6i + b^3*3i))/d - (tan(c/2 + (d*x)/2)^2*(6*a*
b^2 + (3*a^3)/2) + tan(c/2 + (d*x)/2)^6*(102*a*b^2 + 2*a^3) - tan(c/2 + (d*x)/2)^4*(108*a*b^2 + (15*a^3)/4) +
tan(c/2 + (d*x)/2)^3*(26*a^2*b + 8*b^3) + tan(c/2 + (d*x)/2)^7*(30*a^2*b - 8*b^3) - tan(c/2 + (d*x)/2)^5*(58*a
^2*b + 32*b^3) + a^3/4 + 2*a^2*b*tan(c/2 + (d*x)/2))/(d*(16*tan(c/2 + (d*x)/2)^4 - 32*tan(c/2 + (d*x)/2)^6 + 1
6*tan(c/2 + (d*x)/2)^8)) - (tan(c/2 + (d*x)/2)*((15*a^2*b)/8 + b^3/2))/d + (3*a*log(tan(c/2 + (d*x)/2))*(a^2 +
12*b^2))/(8*d) - (a^2*b*tan(c/2 + (d*x)/2)^3)/(8*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \csc ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*csc(c + d*x)**5, x)

________________________________________________________________________________________