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

Optimal. Leaf size=141 \[ -\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {3 a^2 b \csc (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3517, 3768, 3770, 2621, 321, 207, 2622} \[ -\frac {3 a^2 b \csc (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a*b^2*ArcTanh[Cos[c + d*x]])/d + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (
b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^2*b*Csc[c + d*x])/d - (a^3*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (3*a*b^2
*Sec[c + d*x])/d + (b^3*Sec[c + d*x]*Tan[c + d*x])/(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 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 ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \left (a^3 \csc ^3(c+d x)+3 a^2 b \csc ^2(c+d x) \sec (c+d x)+3 a b^2 \csc (c+d x) \sec ^2(c+d x)+b^3 \sec ^3(c+d x)\right ) \, dx\\ &=a^3 \int \csc ^3(c+d x) \, dx+\left (3 a^2 b\right ) \int \csc ^2(c+d x) \sec (c+d x) \, dx+\left (3 a b^2\right ) \int \csc (c+d x) \sec ^2(c+d x) \, dx+b^3 \int \sec ^3(c+d x) \, dx\\ &=-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {b^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^3 \int \csc (c+d x) \, dx+\frac {1}{2} b^3 \int \sec (c+d x) \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x) \tan (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 (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B]  time = 6.19, size = 897, normalized size = 6.36 \[ -\frac {3 a^2 b \cos ^3(c+d x) \tan \left (\frac {1}{2} (c+d x)\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) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {a^3 \cos ^3(c+d x) \csc ^2\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 ^2\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 {3 a^2 b \cos ^3(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^3}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\left (-a^3-6 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}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\left (-b^3-6 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 {\left (a^3+6 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}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\left (b^3+6 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 {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} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*(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) - (3*a^2*b*Cos[c + d*x
]^3*Cot[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a^3*Cos[c + d*x]^3*C
sc[(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 - 6*a*b^2)*Cos[c
+ d*x]^3*Log[Cos[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + ((-6*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) + ((a^3 + 6*a*b^2)*Cos[c + d*x]^3*Log[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) + ((6*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) + (a^3*Cos[c + d*x]^3*Sec[(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) + (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) - (3*a^2*b*Cos[c + d*x]^3*Tan[(c + d*x)/2]*(a + b*Tan[c
+ d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)

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fricas [B]  time = 0.56, size = 299, normalized size = 2.12 \[ -\frac {12 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b^{3} - {\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.11, size = 304, normalized size = 2.16 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, {\left (6 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, {\left (6 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 4 \, {\left (a^{3} + 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) + 4*(6*a^2*b + b^3)*log(abs(tan(1/2*d*x + 1/2*
c) + 1)) - 4*(6*a^2*b + b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 4*(a^3 + 6*a*b^2)*log(abs(tan(1/2*d*x + 1/2*
c))) - (2*a^3*tan(1/2*d*x + 1/2*c)^6 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 8*b
^3*tan(1/2*d*x + 1/2*c)^5 - 3*a^3*tan(1/2*d*x + 1/2*c)^4 + 24*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 24*a^2*b*tan(1/2*
d*x + 1/2*c)^3 - 8*b^3*tan(1/2*d*x + 1/2*c)^3 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*b*tan(1/2*d*x + 1/2*c
) + a^3)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))^2)/d

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maple [A]  time = 0.52, size = 170, normalized size = 1.21 \[ -\frac {a^{3} \cot \left (d x +c \right ) \csc \left (d x +c \right )}{2 d}+\frac {a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {3 a^{2} b}{d \sin \left (d x +c \right )}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 b^{2} a}{d \cos \left (d x +c \right )}+\frac {3 b^{2} a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 171, normalized size = 1.21 \[ \frac {a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a b^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{2} b {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.97, size = 581, normalized size = 4.12 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^3}{2}+24\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^3+24\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^2\,b-4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b+4\,b^3\right )+\frac {a^3}{2}+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-8\,{\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^3}{2}+3\,a\,b^2\right )}{d}-\frac {\mathrm {atan}\left (-\frac {\left (3\,a^2\,b+\frac {b^3}{2}\right )\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {b^3}{2}\right )+6\,a^2\,b+b^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3+6\,a\,b^2\right )\right )\,1{}\mathrm {i}-\left (3\,a^2\,b+\frac {b^3}{2}\right )\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {b^3}{2}\right )-6\,a^2\,b-b^3+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3+6\,a\,b^2\right )\right )\,1{}\mathrm {i}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (36\,a^4\,b^2+12\,a^2\,b^4+b^6\right )-\left (3\,a^2\,b+\frac {b^3}{2}\right )\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {b^3}{2}\right )+6\,a^2\,b+b^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3+6\,a\,b^2\right )\right )-\left (3\,a^2\,b+\frac {b^3}{2}\right )\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b+\frac {b^3}{2}\right )-6\,a^2\,b-b^3+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3+6\,a\,b^2\right )\right )+6\,a\,b^5+6\,a^5\,b+37\,a^3\,b^3}\right )\,\left (a^2\,b\,6{}\mathrm {i}+b^3\,1{}\mathrm {i}\right )}{d}-\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) - (tan(c/2 + (d*x)/2)^4*(24*a*b^2 + a^3/2) - tan(c/2 + (d*x)/2)^2*(24*a*b^2 +
a^3) + tan(c/2 + (d*x)/2)^5*(6*a^2*b - 4*b^3) - tan(c/2 + (d*x)/2)^3*(12*a^2*b + 4*b^3) + a^3/2 + 6*a^2*b*tan
(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 - 8*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6)) + (log(tan(c/2
+ (d*x)/2))*(3*a*b^2 + a^3/2))/d - (atan(-((3*a^2*b + b^3/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + b^3/2) + 6*a^2*
b + b^3 - tan(c/2 + (d*x)/2)*(6*a*b^2 + a^3))*1i - (3*a^2*b + b^3/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + b^3/2) -
6*a^2*b - b^3 + tan(c/2 + (d*x)/2)*(6*a*b^2 + a^3))*1i)/(2*tan(c/2 + (d*x)/2)*(b^6 + 12*a^2*b^4 + 36*a^4*b^2)
- (3*a^2*b + b^3/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + b^3/2) + 6*a^2*b + b^3 - tan(c/2 + (d*x)/2)*(6*a*b^2 + a
^3)) - (3*a^2*b + b^3/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + b^3/2) - 6*a^2*b - b^3 + tan(c/2 + (d*x)/2)*(6*a*b^2
+ a^3)) + 6*a*b^5 + 6*a^5*b + 37*a^3*b^3))*(a^2*b*6i + b^3*1i))/d - (3*a^2*b*tan(c/2 + (d*x)/2))/(2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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