3.48 \(\int \csc ^5(c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=274 \[ \frac{9 a^2 b^2 \sec (c+d x)}{d}-\frac{9 a^2 b^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a^2 b^2 \csc ^2(c+d x) \sec (c+d x)}{d}-\frac{4 a^3 b \csc ^3(c+d x)}{3 d}-\frac{4 a^3 b \csc (c+d x)}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 a^4 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^4 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^4 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{6 a b^3 \csc (c+d x)}{d}+\frac{6 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b^3 \csc (c+d x) \sec ^2(c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}+\frac{b^4 \sec (c+d x)}{d}-\frac{b^4 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

(-3*a^4*ArcTanh[Cos[c + d*x]])/(8*d) - (9*a^2*b^2*ArcTanh[Cos[c + d*x]])/d - (b^4*ArcTanh[Cos[c + d*x]])/d + (
4*a^3*b*ArcTanh[Sin[c + d*x]])/d + (6*a*b^3*ArcTanh[Sin[c + d*x]])/d - (4*a^3*b*Csc[c + d*x])/d - (6*a*b^3*Csc
[c + d*x])/d - (3*a^4*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (4*a^3*b*Csc[c + d*x]^3)/(3*d) - (a^4*Cot[c + d*x]*Cs
c[c + d*x]^3)/(4*d) + (9*a^2*b^2*Sec[c + d*x])/d + (b^4*Sec[c + d*x])/d - (3*a^2*b^2*Csc[c + d*x]^2*Sec[c + d*
x])/d + (2*a*b^3*Csc[c + d*x]*Sec[c + d*x]^2)/d + (b^4*Sec[c + d*x]^3)/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.240883, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 21, 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} \[ \frac{9 a^2 b^2 \sec (c+d x)}{d}-\frac{9 a^2 b^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a^2 b^2 \csc ^2(c+d x) \sec (c+d x)}{d}-\frac{4 a^3 b \csc ^3(c+d x)}{3 d}-\frac{4 a^3 b \csc (c+d x)}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 a^4 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^4 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^4 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{6 a b^3 \csc (c+d x)}{d}+\frac{6 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b^3 \csc (c+d x) \sec ^2(c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}+\frac{b^4 \sec (c+d x)}{d}-\frac{b^4 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^4,x]

[Out]

(-3*a^4*ArcTanh[Cos[c + d*x]])/(8*d) - (9*a^2*b^2*ArcTanh[Cos[c + d*x]])/d - (b^4*ArcTanh[Cos[c + d*x]])/d + (
4*a^3*b*ArcTanh[Sin[c + d*x]])/d + (6*a*b^3*ArcTanh[Sin[c + d*x]])/d - (4*a^3*b*Csc[c + d*x])/d - (6*a*b^3*Csc
[c + d*x])/d - (3*a^4*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (4*a^3*b*Csc[c + d*x]^3)/(3*d) - (a^4*Cot[c + d*x]*Cs
c[c + d*x]^3)/(4*d) + (9*a^2*b^2*Sec[c + d*x])/d + (b^4*Sec[c + d*x])/d - (3*a^2*b^2*Csc[c + d*x]^2*Sec[c + d*
x])/d + (2*a*b^3*Csc[c + d*x]*Sec[c + d*x]^2)/d + (b^4*Sec[c + d*x]^3)/(3*d)

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]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \csc ^5(c+d x) (a+b \tan (c+d x))^4 \, dx &=\int \left (a^4 \csc ^5(c+d x)+4 a^3 b \csc ^4(c+d x) \sec (c+d x)+6 a^2 b^2 \csc ^3(c+d x) \sec ^2(c+d x)+4 a b^3 \csc ^2(c+d x) \sec ^3(c+d x)+b^4 \csc (c+d x) \sec ^4(c+d x)\right ) \, dx\\ &=a^4 \int \csc ^5(c+d x) \, dx+\left (4 a^3 b\right ) \int \csc ^4(c+d x) \sec (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \csc ^2(c+d x) \sec ^3(c+d x) \, dx+b^4 \int \csc (c+d x) \sec ^4(c+d x) \, dx\\ &=-\frac{a^4 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{4} \left (3 a^4\right ) \int \csc ^3(c+d x) \, dx-\frac{\left (4 a^3 b\right ) \operatorname{Subst}\left (\int \frac{x^4}{-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^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{3 a^4 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^4 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^2 b^2 \csc ^2(c+d x) \sec (c+d x)}{d}+\frac{2 a b^3 \csc (c+d x) \sec ^2(c+d x)}{d}+\frac{1}{8} \left (3 a^4\right ) \int \csc (c+d x) \, dx-\frac{\left (4 a^3 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^2 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac{\left (6 a b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac{b^4 \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{3 a^4 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{4 a^3 b \csc (c+d x)}{d}-\frac{6 a b^3 \csc (c+d x)}{d}-\frac{3 a^4 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{4 a^3 b \csc ^3(c+d x)}{3 d}-\frac{a^4 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{9 a^2 b^2 \sec (c+d x)}{d}+\frac{b^4 \sec (c+d x)}{d}-\frac{3 a^2 b^2 \csc ^2(c+d x) \sec (c+d x)}{d}+\frac{2 a b^3 \csc (c+d x) \sec ^2(c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 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 (9 a^2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac{\left (6 a b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{3 a^4 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{9 a^2 b^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{b^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{6 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{4 a^3 b \csc (c+d x)}{d}-\frac{6 a b^3 \csc (c+d x)}{d}-\frac{3 a^4 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{4 a^3 b \csc ^3(c+d x)}{3 d}-\frac{a^4 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{9 a^2 b^2 \sec (c+d x)}{d}+\frac{b^4 \sec (c+d x)}{d}-\frac{3 a^2 b^2 \csc ^2(c+d x) \sec (c+d x)}{d}+\frac{2 a b^3 \csc (c+d x) \sec ^2(c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.2872, size = 1491, normalized size = 5.44 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^4,x]

[Out]

(b^2*(36*a^2 + 7*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) + ((-7*
a^3*b*Cos[(c + d*x)/2] - 6*a*b^3*Cos[(c + d*x)/2])*Cos[c + d*x]^4*Csc[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(3*
d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (3*(a^4 + 8*a^2*b^2)*Cos[c + d*x]^4*Csc[(c + d*x)/2]^2*(a + b*Tan[c +
d*x])^4)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (a^3*b*Cos[c + d*x]^4*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]
^2*(a + b*Tan[c + d*x])^4)/(6*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (a^4*Cos[c + d*x]^4*Csc[(c + d*x)/2]^4*
(a + b*Tan[c + d*x])^4)/(64*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-3*a^4 - 72*a^2*b^2 - 8*b^4)*Cos[c + d*
x]^4*Log[Cos[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*(2*a^3*b + 3
*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) + ((3*a^4 + 72*a^2*b^2 + 8*b^4)*Cos[c + d*x]^4*Log[Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)
/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*(2*a^3*b + 3*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) + (3*(a^4 + 8*a^2*b^2)*Cos[c + d
*x]^4*Sec[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^4)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (a^4*Cos[c + d*x
]^4*Sec[(c + d*x)/2]^4*(a + b*Tan[c + d*x])^4)/(64*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*Sec[(c + d*x)/2]*(-7*a^3*b*Sin[(c + d*x)/2] - 6*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c +
d*x])^4)/(3*d*(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] - 7*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] + 7*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) - (a^3*b*Cos[c + d*x]^4*Sec[(c + d
*x)/2]^2*Tan[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(6*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)

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Maple [A]  time = 0.082, size = 317, normalized size = 1.2 \begin{align*}{\frac{{b}^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{b}^{4}}{d\cos \left ( dx+c \right ) }}+{\frac{{b}^{4}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{{b}^{3}a}{d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-6\,{\frac{{b}^{3}a}{d\sin \left ( dx+c \right ) }}+6\,{\frac{{b}^{3}a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{{a}^{2}{b}^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+9\,{\frac{{a}^{2}{b}^{2}}{d\cos \left ( dx+c \right ) }}+9\,{\frac{{a}^{2}{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{4\,b{a}^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-4\,{\frac{b{a}^{3}}{d\sin \left ( dx+c \right ) }}+4\,{\frac{b{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{4}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{4}\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{4}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+b*tan(d*x+c))^4,x)

[Out]

1/3/d*b^4/cos(d*x+c)^3+1/d*b^4/cos(d*x+c)+1/d*b^4*ln(csc(d*x+c)-cot(d*x+c))+2/d*b^3*a/sin(d*x+c)/cos(d*x+c)^2-
6/d*b^3*a/sin(d*x+c)+6/d*b^3*a*ln(sec(d*x+c)+tan(d*x+c))-3/d*a^2*b^2/sin(d*x+c)^2/cos(d*x+c)+9/d*a^2*b^2/cos(d
*x+c)+9/d*a^2*b^2*ln(csc(d*x+c)-cot(d*x+c))-4/3/d*b*a^3/sin(d*x+c)^3-4/d*b*a^3/sin(d*x+c)+4/d*b*a^3*ln(sec(d*x
+c)+tan(d*x+c))-1/4*a^4*cot(d*x+c)*csc(d*x+c)^3/d-3/8*a^4*cot(d*x+c)*csc(d*x+c)/d+3/8/d*a^4*ln(csc(d*x+c)-cot(
d*x+c))

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Maxima [A]  time = 1.17311, size = 410, normalized size = 1.5 \begin{align*} \frac{3 \, a^{4}{\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 )} + 72 \, a^{2} 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 )} - 48 \, a 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 \, b^{4}{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 32 \, a^{3} 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 )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/48*(3*a^4*(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)) + 72*a^2*b^2*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3*l
og(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 48*a*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*b^4*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 -
3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 32*a^3*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

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Fricas [B]  time = 3.98277, size = 1289, normalized size = 4.7 \begin{align*} \frac{6 \,{\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{6} - 10 \,{\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 16 \, b^{4} + 16 \,{\left (18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left ({\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{7} - 2 \,{\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\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 (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{7} - 2 \,{\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (3 \, a^{4} + 72 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 48 \,{\left ({\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \,{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} +{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \,{\left ({\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \,{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} +{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 32 \,{\left (3 \,{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 3 \, a b^{3} \cos \left (d x + c\right ) - 4 \,{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right )^{7} - 2 \, d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/48*(6*(3*a^4 + 72*a^2*b^2 + 8*b^4)*cos(d*x + c)^6 - 10*(3*a^4 + 72*a^2*b^2 + 8*b^4)*cos(d*x + c)^4 + 16*b^4
+ 16*(18*a^2*b^2 + b^4)*cos(d*x + c)^2 - 3*((3*a^4 + 72*a^2*b^2 + 8*b^4)*cos(d*x + c)^7 - 2*(3*a^4 + 72*a^2*b^
2 + 8*b^4)*cos(d*x + c)^5 + (3*a^4 + 72*a^2*b^2 + 8*b^4)*cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) + 3*((3*a
^4 + 72*a^2*b^2 + 8*b^4)*cos(d*x + c)^7 - 2*(3*a^4 + 72*a^2*b^2 + 8*b^4)*cos(d*x + c)^5 + (3*a^4 + 72*a^2*b^2
+ 8*b^4)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) + 48*((2*a^3*b + 3*a*b^3)*cos(d*x + c)^7 - 2*(2*a^3*b +
3*a*b^3)*cos(d*x + c)^5 + (2*a^3*b + 3*a*b^3)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 48*((2*a^3*b + 3*a*b^3)*
cos(d*x + c)^7 - 2*(2*a^3*b + 3*a*b^3)*cos(d*x + c)^5 + (2*a^3*b + 3*a*b^3)*cos(d*x + c)^3)*log(-sin(d*x + c)
+ 1) + 32*(3*(2*a^3*b + 3*a*b^3)*cos(d*x + c)^5 + 3*a*b^3*cos(d*x + c) - 4*(2*a^3*b + 3*a*b^3)*cos(d*x + c)^3)
*sin(d*x + c))/(d*cos(d*x + c)^7 - 2*d*cos(d*x + c)^5 + d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5*(a+b*tan(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 2.68359, size = 647, normalized size = 2.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/192*(3*a^4*tan(1/2*d*x + 1/2*c)^4 - 32*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 24*a^4*tan(1/2*d*x + 1/2*c)^2 + 144*a^
2*b^2*tan(1/2*d*x + 1/2*c)^2 - 480*a^3*b*tan(1/2*d*x + 1/2*c) - 384*a*b^3*tan(1/2*d*x + 1/2*c) + 384*(2*a^3*b
+ 3*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 384*(2*a^3*b + 3*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2
4*(3*a^4 + 72*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c))) + 256*(3*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 9*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 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 3*b^4*tan(1/2*d*
x + 1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c) - 9*a^2*b^2 - 2*b^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3 - (150*a^4*tan
(1/2*d*x + 1/2*c)^4 + 3600*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 400*b^4*tan(1/2*d*x + 1/2*c)^4 + 480*a^3*b*tan(1/2
*d*x + 1/2*c)^3 + 384*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 24*a^4*tan(1/2*d*x + 1/2*c)^2 + 144*a^2*b^2*tan(1/2*d*x +
1/2*c)^2 + 32*a^3*b*tan(1/2*d*x + 1/2*c) + 3*a^4)/tan(1/2*d*x + 1/2*c)^4)/d