3.46 \(\int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=161 \[ -\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {4 a^3 b \csc (c+d x)}{d}+\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {6 a^2 b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {2 a b^3 \tan (c+d x) \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d} \]

[Out]

-1/2*a^4*arctanh(cos(d*x+c))/d-6*a^2*b^2*arctanh(cos(d*x+c))/d+4*a^3*b*arctanh(sin(d*x+c))/d+2*a*b^3*arctanh(s
in(d*x+c))/d-4*a^3*b*csc(d*x+c)/d-1/2*a^4*cot(d*x+c)*csc(d*x+c)/d+6*a^2*b^2*sec(d*x+c)/d+1/3*b^4*sec(d*x+c)^3/
d+2*a*b^3*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.16, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3517, 3768, 3770, 2621, 321, 207, 2622, 2606, 30} \[ \frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {6 a^2 b^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {4 a^3 b \csc (c+d x)}{d}+\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {2 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {2 a b^3 \tan (c+d x) \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]

[Out]

-(a^4*ArcTanh[Cos[c + d*x]])/(2*d) - (6*a^2*b^2*ArcTanh[Cos[c + d*x]])/d + (4*a^3*b*ArcTanh[Sin[c + d*x]])/d +
(2*a*b^3*ArcTanh[Sin[c + d*x]])/d - (4*a^3*b*Csc[c + d*x])/d - (a^4*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (6*a^2
*b^2*Sec[c + d*x])/d + (b^4*Sec[c + d*x]^3)/(3*d) + (2*a*b^3*Sec[c + d*x]*Tan[c + d*x])/d

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 2606

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

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 ^3(c+d x) (a+b \tan (c+d x))^4 \, dx &=\int \left (a^4 \csc ^3(c+d x)+4 a^3 b \csc ^2(c+d x) \sec (c+d x)+6 a^2 b^2 \csc (c+d x) \sec ^2(c+d x)+4 a b^3 \sec ^3(c+d x)+b^4 \sec ^3(c+d x) \tan (c+d x)\right ) \, dx\\ &=a^4 \int \csc ^3(c+d x) \, dx+\left (4 a^3 b\right ) \int \csc ^2(c+d x) \sec (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \csc (c+d x) \sec ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \sec ^3(c+d x) \, dx+b^4 \int \sec ^3(c+d x) \tan (c+d x) \, dx\\ &=-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d}+\frac {1}{2} a^4 \int \csc (c+d x) \, dx+\left (2 a b^3\right ) \int \sec (c+d x) \, dx-\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (6 a^2 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}+\frac {b^4 \operatorname {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {2 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {4 a^3 b \csc (c+d x)}{d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d}-\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (6 a^2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {6 a^2 b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {2 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {4 a^3 b \csc (c+d x)}{d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 6.21, size = 1128, normalized size = 7.01 \[ -\frac {2 a^3 b \cos ^4(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {b^2 \left (36 a^2+b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{6 d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {a^4 \cos ^4(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{8 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {a^4 \cos ^4(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{8 d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {2 a^3 b \cos ^4(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (-a^4-12 b^2 a^2\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{2 d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {2 \left (2 b a^3+b^3 a\right ) \cos ^4(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))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (a^4+12 b^2 a^2\right ) \cos ^4(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{2 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {2 \left (2 b a^3+b^3 a\right ) \cos ^4(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))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {b^4 \cos ^4(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\cos ^4(c+d x) \left (-\sin \left (\frac {1}{2} (c+d x)\right ) b^4-36 a^2 \sin \left (\frac {1}{2} (c+d x)\right ) b^2\right ) (a+b \tan (c+d x))^4}{6 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))^4}+\frac {\cos ^4(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right ) b^4+36 a^2 \sin \left (\frac {1}{2} (c+d x)\right ) b^2\right ) (a+b \tan (c+d x))^4}{6 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))^4}+\frac {\left (b^4+12 a b^3\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{12 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))^4}+\frac {\left (b^4-12 a b^3\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{12 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))^4}-\frac {b^4 \cos ^4(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]

[Out]

(b^2*(36*a^2 + b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(6*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*a^3*
b*Cos[c + d*x]^4*Cot[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (a^4*Cos[c
+ d*x]^4*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^4)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-a^4 - 12*a
^2*b^2)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)
- (2*(2*a^3*b + a*b^3)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(d*(a*
Cos[c + d*x] + b*Sin[c + d*x])^4) + ((a^4 + 12*a^2*b^2)*Cos[c + d*x]^4*Log[Sin[(c + d*x)/2]]*(a + b*Tan[c + d*
x])^4)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*(2*a^3*b + a*b^3)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (a^4*Cos[c + d*x]^4*Sec[(c
+ d*x)/2]^2*(a + b*Tan[c + d*x])^4)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((12*a*b^3 + b^4)*Cos[c + d*x
]^4*(a + b*Tan[c + d*x])^4)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)
+ (b^4*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(
a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (b^4*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(
c + d*x)/2] + Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-12*a*b^3 + b^4)*Cos[c + d*x]^4*(a
+ b*Tan[c + d*x])^4)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos
[c + d*x]^4*(-36*a^2*b^2*Sin[(c + d*x)/2] - b^4*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(36*a^2*b^2*Sin[(c + d*x)/2] + b
^4*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Si
n[c + d*x])^4) - (2*a^3*b*Cos[c + d*x]^4*Tan[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c
+ d*x])^4)

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fricas [B]  time = 0.55, size = 346, normalized size = 2.15 \[ \frac {6 \, {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, b^{4} - 4 \, {\left (18 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, {\left ({\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left ({\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 24 \, {\left (a b^{3} \cos \left (d x + c\right ) - {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{5} - d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/12*(6*(a^4 + 12*a^2*b^2)*cos(d*x + c)^4 - 4*b^4 - 4*(18*a^2*b^2 - b^4)*cos(d*x + c)^2 - 3*((a^4 + 12*a^2*b^2
)*cos(d*x + c)^5 - (a^4 + 12*a^2*b^2)*cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) + 3*((a^4 + 12*a^2*b^2)*cos(
d*x + c)^5 - (a^4 + 12*a^2*b^2)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) + 12*((2*a^3*b + a*b^3)*cos(d*x +
c)^5 - (2*a^3*b + a*b^3)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 12*((2*a^3*b + a*b^3)*cos(d*x + c)^5 - (2*a^
3*b + a*b^3)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 24*(a*b^3*cos(d*x + c) - (2*a^3*b + a*b^3)*cos(d*x + c)^
3)*sin(d*x + c))/(d*cos(d*x + c)^5 - d*cos(d*x + c)^3)

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giac [A]  time = 12.82, size = 300, normalized size = 1.86 \[ \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, {\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, {\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 12 \, {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {3 \, {\left (6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{4}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {16 \, {\left (6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b^{2} - b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/24*(3*a^4*tan(1/2*d*x + 1/2*c)^2 - 48*a^3*b*tan(1/2*d*x + 1/2*c) + 48*(2*a^3*b + a*b^3)*log(abs(tan(1/2*d*x
+ 1/2*c) + 1)) - 48*(2*a^3*b + a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 12*(a^4 + 12*a^2*b^2)*log(abs(tan(1
/2*d*x + 1/2*c))) - 3*(6*a^4*tan(1/2*d*x + 1/2*c)^2 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 16*a^3*b*tan(1/2*d*x
+ 1/2*c) + a^4)/tan(1/2*d*x + 1/2*c)^2 + 16*(6*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b^2*tan(1/2*d*x + 1/2*c)
^4 - 3*b^4*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a*b^3*tan(1/2*d*x + 1/2*c) - 18*a^2*
b^2 - b^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 0.52, size = 192, normalized size = 1.19 \[ -\frac {a^{4} \cot \left (d x +c \right ) \csc \left (d x +c \right )}{2 d}+\frac {a^{4} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {4 a^{3} b}{d \sin \left (d x +c \right )}+\frac {4 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2}}{d \cos \left (d x +c \right )}+\frac {6 a^{2} b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {2 a \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {2 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{4}}{3 d \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x)

[Out]

-1/2*a^4*cot(d*x+c)*csc(d*x+c)/d+1/2/d*a^4*ln(csc(d*x+c)-cot(d*x+c))-4/d*a^3*b/sin(d*x+c)+4/d*a^3*b*ln(sec(d*x
+c)+tan(d*x+c))+6/d*a^2*b^2/cos(d*x+c)+6/d*a^2*b^2*ln(csc(d*x+c)-cot(d*x+c))+2*a*b^3*sec(d*x+c)*tan(d*x+c)/d+2
/d*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/3/d*b^4/cos(d*x+c)^3

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maxima [A]  time = 0.48, size = 188, normalized size = 1.17 \[ \frac {3 \, a^{4} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a^{2} b^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 24 \, a^{3} b {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {4 \, b^{4}}{\cos \left (d x + c\right )^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/12*(3*a^4*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 12*a*b^3*(
2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*a^2*b^2*(2/cos(d*x +
c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 24*a^3*b*(2/sin(d*x + c) - log(sin(d*x + c) + 1) + log(
sin(d*x + c) - 1)) + 4*b^4/cos(d*x + c)^3)/d

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mupad [B]  time = 4.07, size = 670, normalized size = 4.16 \[ \frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {\frac {a^4}{2}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a^4}{2}+48\,a^2\,b^2+\frac {8\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^4}{2}+48\,a^2\,b^2+8\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {3\,a^4}{2}+96\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (16\,a\,b^3-8\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (24\,a^3\,b+16\,a\,b^3\right )+8\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+24\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{d\,\left (-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^4}{2}+6\,a^2\,b^2\right )}{d}-\frac {2\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {a\,b\,\mathrm {atan}\left (\frac {a\,b\,\left (2\,a^2+b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+12\,a^2\,b^2\right )-4\,a\,b^3-8\,a^3\,b+12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )\right )\,2{}\mathrm {i}-a\,b\,\left (2\,a^2+b^2\right )\,\left (4\,a\,b^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+12\,a^2\,b^2\right )+8\,a^3\,b+12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )\right )\,2{}\mathrm {i}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (64\,a^6\,b^2+64\,a^4\,b^4+16\,a^2\,b^6\right )+8\,a^7\,b+48\,a^3\,b^5+100\,a^5\,b^3-2\,a\,b\,\left (2\,a^2+b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+12\,a^2\,b^2\right )-4\,a\,b^3-8\,a^3\,b+12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )\right )-2\,a\,b\,\left (2\,a^2+b^2\right )\,\left (4\,a\,b^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+12\,a^2\,b^2\right )+8\,a^3\,b+12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )\right )}\right )\,\left (2\,a^2+b^2\right )\,4{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^4/sin(c + d*x)^3,x)

[Out]

(a^4*tan(c/2 + (d*x)/2)^2)/(8*d) - (a^4/2 - tan(c/2 + (d*x)/2)^2*((3*a^4)/2 + (8*b^4)/3 + 48*a^2*b^2) - tan(c/
2 + (d*x)/2)^6*(a^4/2 + 8*b^4 + 48*a^2*b^2) + tan(c/2 + (d*x)/2)^4*((3*a^4)/2 + 96*a^2*b^2) + tan(c/2 + (d*x)/
2)^7*(16*a*b^3 - 8*a^3*b) - tan(c/2 + (d*x)/2)^3*(16*a*b^3 + 24*a^3*b) + 8*a^3*b*tan(c/2 + (d*x)/2) + 24*a^3*b
*tan(c/2 + (d*x)/2)^5)/(d*(4*tan(c/2 + (d*x)/2)^2 - 12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)^6 - 4*tan(
c/2 + (d*x)/2)^8)) + (log(tan(c/2 + (d*x)/2))*(a^4/2 + 6*a^2*b^2))/d - (2*a^3*b*tan(c/2 + (d*x)/2))/d - (a*b*a
tan((a*b*(2*a^2 + b^2)*(tan(c/2 + (d*x)/2)*(a^4 + 12*a^2*b^2) - 4*a*b^3 - 8*a^3*b + 12*a*b*tan(c/2 + (d*x)/2)*
(2*a^2 + b^2))*2i - a*b*(2*a^2 + b^2)*(4*a*b^3 - tan(c/2 + (d*x)/2)*(a^4 + 12*a^2*b^2) + 8*a^3*b + 12*a*b*tan(
c/2 + (d*x)/2)*(2*a^2 + b^2))*2i)/(2*tan(c/2 + (d*x)/2)*(16*a^2*b^6 + 64*a^4*b^4 + 64*a^6*b^2) + 8*a^7*b + 48*
a^3*b^5 + 100*a^5*b^3 - 2*a*b*(2*a^2 + b^2)*(tan(c/2 + (d*x)/2)*(a^4 + 12*a^2*b^2) - 4*a*b^3 - 8*a^3*b + 12*a*
b*tan(c/2 + (d*x)/2)*(2*a^2 + b^2)) - 2*a*b*(2*a^2 + b^2)*(4*a*b^3 - tan(c/2 + (d*x)/2)*(a^4 + 12*a^2*b^2) + 8
*a^3*b + 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 + b^2))))*(2*a^2 + b^2)*4i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \csc ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3*(a+b*tan(d*x+c))**4,x)

[Out]

Integral((a + b*tan(c + d*x))**4*csc(c + d*x)**3, x)

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