\(\int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\) [30]

Optimal result

Integrand size = 21, antiderivative size = 165 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {3 b^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \csc (c+d x)}{d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {2 a b \csc ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {3 b^2 \sec (c+d x)}{2 d}-\frac {b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3598, 3853, 3855, 2701, 308, 213, 2702, 294, 327} \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \csc ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}-\frac {3 b^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {3 b^2 \sec (c+d x)}{2 d}-\frac {b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

[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 213

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

Rule 294

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 308

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 327

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 2701

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 2702

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 3598

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 3853

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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) \text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {b^2 \text {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) \text {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 ) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d} \\ & = -\frac {3 a^2 \text {arctanh}(\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) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d} \\ & = -\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {3 b^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {2 a b \text {arctanh}(\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(994\) vs. \(2(165)=330\).

Time = 7.25 (sec) , antiderivative size = 994, normalized size of antiderivative = 6.02 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=\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 {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 {\left (-3 a^2-4 b^2\right ) \cos ^2(c+d x) \csc ^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 {a b \cos ^2(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\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) \csc ^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 {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 {\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 {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) \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}-\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} \]

[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)

Maple [A] (verified)

Time = 6.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (\left (-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\)	\(145\)
default	\(\frac {b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (\left (-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\)	\(145\)
risch	\(\frac {{\mathrm e}^{i \left (d x +c \right )} \left (9 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+36 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-48 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}-24 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-96 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+160 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}-66 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+120 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-96 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-160 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a^{2}+36 b^{2}+48 i a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{2 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{2 d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(358\)

[In]

int(csc(d*x+c)^5*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (151) = 302\).

Time = 0.30 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.02 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {18 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 30 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, b^{2} - 9 \, {\left ({\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left ({\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 48 \, {\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \, {\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 32 \, {\left (3 \, a b \cos \left (d x + c\right )^{3} - 4 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{5} - 2 \, d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )\right )}} \]

[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))

Sympy [F]

\[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \csc ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**5*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*csc(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.13 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, b^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 16 \, a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.63 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 384 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 384 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, {\left (a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {384 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 4.22 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.29 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{8}+\frac {3\,b^2}{2}\right )}{d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {7\,a^2}{4}+2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2+34\,b^2\right )+\frac {a^2}{4}+\frac {56\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-20\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{8}+\frac {b^2}{8}\right )}{d}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {12\,a\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^3\,b-16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2+12\,a\,b^3}-\frac {16\,a^2\,b^2}{3\,a^3\,b-16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2+12\,a\,b^3}+\frac {3\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^3\,b-16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2+12\,a\,b^3}\right )}{d}-\frac {5\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d} \]

[In]

int((a + b*tan(c + d*x))^2/sin(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2))*((3*a^2)/8 + (3*b^2)/2))/d + (a^2*tan(c/2 + (d*x)/2)^4)/(64*d) - (tan(c/2 + (d*x)/2)^
2*((7*a^2)/4 + 2*b^2) - tan(c/2 + (d*x)/2)^4*(2*a^2 + 34*b^2) + a^2/4 + (56*a*b*tan(c/2 + (d*x)/2)^3)/3 - 20*a
*b*tan(c/2 + (d*x)/2)^5 + (4*a*b*tan(c/2 + (d*x)/2))/3)/(d*(16*tan(c/2 + (d*x)/2)^4 - 16*tan(c/2 + (d*x)/2)^6)
) + (tan(c/2 + (d*x)/2)^2*(a^2/8 + b^2/8))/d - (a*b*tan(c/2 + (d*x)/2)^3)/(12*d) + (4*a*b*atanh((12*a*b^3*tan(
c/2 + (d*x)/2))/(12*a*b^3 + 3*a^3*b - 16*a^2*b^2*tan(c/2 + (d*x)/2)) - (16*a^2*b^2)/(12*a*b^3 + 3*a^3*b - 16*a
^2*b^2*tan(c/2 + (d*x)/2)) + (3*a^3*b*tan(c/2 + (d*x)/2))/(12*a*b^3 + 3*a^3*b - 16*a^2*b^2*tan(c/2 + (d*x)/2))
))/d - (5*a*b*tan(c/2 + (d*x)/2))/(4*d)