3.30 \(\int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=165 \[ -\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 b^2 \sec (c+d x)}{2 d}-\frac{3 b^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.160082, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 13, 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} \[ -\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 b^2 \sec (c+d x)}{2 d}-\frac{3 b^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(-3*a^2*ArcTanh[Cos[c + d*x]])/(8*d) - (3*b^2*ArcTanh[Cos[c + d*x]])/(2*d) + (2*a*b*ArcTanh[Sin[c + d*x]])/d -
(2*a*b*Csc[c + d*x])/d - (3*a^2*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (2*a*b*Csc[c + d*x]^3)/(3*d) - (a^2*Cot[c
+ d*x]*Csc[c + d*x]^3)/(4*d) + (3*b^2*Sec[c + d*x])/(2*d) - (b^2*Csc[c + d*x]^2*Sec[c + d*x])/(2*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))^2 \, dx &=\int \left (a^2 \csc ^5(c+d x)+2 a b \csc ^4(c+d x) \sec (c+d x)+b^2 \csc ^3(c+d x) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 \int \csc ^5(c+d x) \, dx+(2 a b) \int \csc ^4(c+d x) \sec (c+d x) \, dx+b^2 \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx\\ &=-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{4} \left (3 a^2\right ) \int \csc ^3(c+d x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{1}{8} \left (3 a^2\right ) \int \csc (c+d x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{2 a b \csc (c+d x)}{d}-\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{3 b^2 \sec (c+d x)}{2 d}-\frac{b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{3 b^2 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a b \csc (c+d x)}{d}-\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{3 b^2 \sec (c+d x)}{2 d}-\frac{b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 6.19963, size = 994, normalized size = 6.02 \[ -\frac{a^2 \cos ^2(c+d x) (a+b \tan (c+d x))^2 \csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{\left (-3 a^2-4 b^2\right ) \cos ^2(c+d x) (a+b \tan (c+d x))^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{a b \cos ^2(c+d x) \cot \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{12 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{7 a b \cos ^2(c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2}{6 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{a b \cos ^2(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))^2}{12 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{a^2 \cos ^2(c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2}{64 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{b^2 \cos ^2(c+d x) (a+b \tan (c+d x))^2}{d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{\left (3 a^2+4 b^2\right ) \cos ^2(c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2}{32 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{7 a b \cos ^2(c+d x) \cot \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2}{6 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{3 \left (a^2+4 b^2\right ) \cos ^2(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}{8 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{2 a b \cos ^2(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))^2}{d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{3 \left (a^2+4 b^2\right ) \cos ^2(c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}{8 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{2 a b \cos ^2(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))^2}{d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{b^2 \cos ^2(c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2}{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))^2}-\frac{b^2 \cos ^2(c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^2}{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))^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(b^2*Cos[c + d*x]^2*(a + b*Tan[c + d*x])^2)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (7*a*b*Cos[c + d*x]^2*Co
t[(c + d*x)/2]*(a + b*Tan[c + d*x])^2)/(6*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + ((-3*a^2 - 4*b^2)*Cos[c + d
*x]^2*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^2)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (a*b*Cos[c + d*x
]^2*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^2)/(12*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (
a^2*Cos[c + d*x]^2*Csc[(c + d*x)/2]^4*(a + b*Tan[c + d*x])^2)/(64*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (3*
(a^2 + 4*b^2)*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2]]*(a + b*Tan[c + d*x])^2)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*
x])^2) - (2*a*b*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^2)/(d*(a*Cos[c +
d*x] + b*Sin[c + d*x])^2) + (3*(a^2 + 4*b^2)*Cos[c + d*x]^2*Log[Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^2)/(8*d
*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + (2*a*b*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*
Tan[c + d*x])^2)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + ((3*a^2 + 4*b^2)*Cos[c + d*x]^2*Sec[(c + d*x)/2]^2*
(a + b*Tan[c + d*x])^2)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + (a^2*Cos[c + d*x]^2*Sec[(c + d*x)/2]^4*(a
+ b*Tan[c + d*x])^2)/(64*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + (b^2*Cos[c + d*x]^2*Sin[(c + d*x)/2]*(a + b
*Tan[c + d*x])^2)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (b^2*Cos[c +
d*x]^2*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^2)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*
Sin[c + d*x])^2) - (7*a*b*Cos[c + d*x]^2*Tan[(c + d*x)/2]*(a + b*Tan[c + d*x])^2)/(6*d*(a*Cos[c + d*x] + b*Sin
[c + d*x])^2) - (a*b*Cos[c + d*x]^2*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]*(a + b*Tan[c + d*x])^2)/(12*d*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^2)

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Maple [A]  time = 0.054, size = 183, normalized size = 1.1 \begin{align*} -{\frac{{b}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+{\frac{3\,{b}^{2}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{3\,{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{2\,ab}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{ab}{d\sin \left ( dx+c \right ) }}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{2}\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}\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))^2,x)

[Out]

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

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Maxima [A]  time = 1.07451, size = 252, normalized size = 1.53 \begin{align*} \frac{3 \, a^{2}{\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 \, 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 )} - 16 \, a 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))^2,x, algorithm="maxima")

[Out]

1/48*(3*a^2*(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*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)) - 16*a*b*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x
+ c) + 1) + 3*log(sin(d*x + c) - 1)))/d

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Fricas [B]  time = 2.97415, size = 856, normalized size = 5.19 \begin{align*} \frac{18 \,{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 30 \,{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, b^{2} - 9 \,{\left ({\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 2 \,{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 9 \,{\left ({\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 2 \,{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 48 \,{\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \,{\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 32 \,{\left (3 \, a b \cos \left (d x + c\right )^{3} - 4 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right )^{5} - 2 \, d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )\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))^2,x, algorithm="fricas")

[Out]

1/48*(18*(a^2 + 4*b^2)*cos(d*x + c)^4 - 30*(a^2 + 4*b^2)*cos(d*x + c)^2 + 48*b^2 - 9*((a^2 + 4*b^2)*cos(d*x +
c)^5 - 2*(a^2 + 4*b^2)*cos(d*x + c)^3 + (a^2 + 4*b^2)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 9*((a^2 + 4*
b^2)*cos(d*x + c)^5 - 2*(a^2 + 4*b^2)*cos(d*x + c)^3 + (a^2 + 4*b^2)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2
) + 48*(a*b*cos(d*x + c)^5 - 2*a*b*cos(d*x + c)^3 + a*b*cos(d*x + c))*log(sin(d*x + c) + 1) - 48*(a*b*cos(d*x
+ c)^5 - 2*a*b*cos(d*x + c)^3 + a*b*cos(d*x + c))*log(-sin(d*x + c) + 1) + 32*(3*a*b*cos(d*x + c)^3 - 4*a*b*co
s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5 - 2*d*cos(d*x + c)^3 + d*cos(d*x + c))

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.6524, size = 363, normalized size = 2.2 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 384 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 384 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 72 \,{\left (a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{384 \, b^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \frac{150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 600 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, 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))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*a*b*tan(1/2*d*x + 1/2*c)^3 + 24*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*t
an(1/2*d*x + 1/2*c)^2 + 384*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 384*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)
) - 240*a*b*tan(1/2*d*x + 1/2*c) + 72*(a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - 384*b^2/(tan(1/2*d*x + 1/
2*c)^2 - 1) - (150*a^2*tan(1/2*d*x + 1/2*c)^4 + 600*b^2*tan(1/2*d*x + 1/2*c)^4 + 240*a*b*tan(1/2*d*x + 1/2*c)^
3 + 24*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*tan(1/2*d*x + 1/2*c)^2 + 16*a*b*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1
/2*d*x + 1/2*c)^4)/d