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

Optimal. Leaf size=205 \[ -\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{3 a^2 b \csc (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{a b^2 \cot ^3(c+d x)}{d}-\frac{6 a b^2 \cot (c+d x)}{d}-\frac{5 b^3 \csc ^3(c+d x)}{6 d}-\frac{5 b^3 \csc (c+d x)}{2 d}+\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \csc ^3(c+d x) \sec ^2(c+d x)}{2 d} \]

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (5*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (6*a*b^2*Cot[
c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) - (a*b^2*Cot[c + d*x]^3)/d - (3*a^2*b*Csc[c + d*x])/d - (5*b^3*Csc[c
+ d*x])/(2*d) - (a^2*b*Csc[c + d*x]^3)/d - (5*b^3*Csc[c + d*x]^3)/(6*d) + (b^3*Csc[c + d*x]^3*Sec[c + d*x]^2)/
(2*d) + (3*a*b^2*Tan[c + d*x])/d

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Rubi [A]  time = 0.291242, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3872, 2912, 3767, 2621, 302, 207, 2620, 270, 288} \[ -\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{3 a^2 b \csc (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{a b^2 \cot ^3(c+d x)}{d}-\frac{6 a b^2 \cot (c+d x)}{d}-\frac{5 b^3 \csc ^3(c+d x)}{6 d}-\frac{5 b^3 \csc (c+d x)}{2 d}+\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \csc ^3(c+d x) \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (5*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (6*a*b^2*Cot[
c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) - (a*b^2*Cot[c + d*x]^3)/d - (3*a^2*b*Csc[c + d*x])/d - (5*b^3*Csc[c
+ d*x])/(2*d) - (a^2*b*Csc[c + d*x]^3)/d - (5*b^3*Csc[c + d*x]^3)/(6*d) + (b^3*Csc[c + d*x]^3*Sec[c + d*x]^2)/
(2*d) + (3*a*b^2*Tan[c + d*x])/d

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2912

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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]

Rubi steps

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

Mathematica [B]  time = 0.915758, size = 610, normalized size = 2.98 \[ -\frac{\csc ^7\left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (32 a \left (a^2+3 b^2\right ) \cos (c+d x)+8 \left (6 a^2 b+5 b^3\right ) \cos (2 (c+d x))-36 a^2 b \cos (4 (c+d x))+36 a^2 b \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-36 a^2 b \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+18 a^2 b \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-18 a^2 b \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-18 a^2 b \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+18 a^2 b \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+84 a^2 b+4 a^3 \cos (3 (c+d x))-4 a^3 \cos (5 (c+d x))+48 a b^2 \cos (3 (c+d x))-48 a b^2 \cos (5 (c+d x))-30 b^3 \cos (4 (c+d x))+30 b^3 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-30 b^3 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+15 b^3 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-15 b^3 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-15 b^3 \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 b^3 \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+22 b^3\right )}{768 d \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]^7*Sec[(c + d*x)/2]^3*(84*a^2*b + 22*b^3 + 32*a*(a^2 + 3*b^2)*Cos[c + d*x] + 8*(6*a^2*b + 5*
b^3)*Cos[2*(c + d*x)] + 4*a^3*Cos[3*(c + d*x)] + 48*a*b^2*Cos[3*(c + d*x)] - 36*a^2*b*Cos[4*(c + d*x)] - 30*b^
3*Cos[4*(c + d*x)] - 4*a^3*Cos[5*(c + d*x)] - 48*a*b^2*Cos[5*(c + d*x)] + 36*a^2*b*Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]]*Sin[c + d*x] + 30*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 36*a^2*b*Log[Cos[(
c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 30*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] +
18*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 15*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]]*Sin[3*(c + d*x)] - 18*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 15*b^3*Log[Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 18*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[5*(c
+ d*x)] - 15*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 18*a^2*b*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 15*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(768*d
*(-1 + Cot[(c + d*x)/2]^2)^2)

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Maple [A]  time = 0.05, size = 246, normalized size = 1.2 \begin{align*} -{\frac{2\,{a}^{3}\cot \left ( dx+c \right ) }{3\,d}}-{\frac{{a}^{3}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{{a}^{2}b}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-3\,{\frac{{a}^{2}b}{d\sin \left ( dx+c \right ) }}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{a{b}^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+4\,{\frac{a{b}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-8\,{\frac{a{b}^{2}\cot \left ( dx+c \right ) }{d}}-{\frac{{b}^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,{b}^{3}}{6\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,{b}^{3}}{2\,d\sin \left ( dx+c \right ) }}+{\frac{5\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^3*cot(d*x+c)/d-1/3/d*a^3*cot(d*x+c)*csc(d*x+c)^2-1/d*a^2*b/sin(d*x+c)^3-3/d*a^2*b/sin(d*x+c)+3/d*a^2*b*
ln(sec(d*x+c)+tan(d*x+c))-1/d*a*b^2/sin(d*x+c)^3/cos(d*x+c)+4/d*a*b^2/sin(d*x+c)/cos(d*x+c)-8*a*b^2*cot(d*x+c)
/d-1/3/d*b^3/sin(d*x+c)^3/cos(d*x+c)^2+5/6/d*b^3/sin(d*x+c)/cos(d*x+c)^2-5/2/d*b^3/sin(d*x+c)+5/2/d*b^3*ln(sec
(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.03308, size = 257, normalized size = 1.25 \begin{align*} -\frac{b^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{3}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} 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 )} + 12 \, a b^{2}{\left (\frac{6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac{4 \,{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{3}}{\tan \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(b^3*(2*(15*sin(d*x + c)^4 - 10*sin(d*x + c)^2 - 2)/(sin(d*x + c)^5 - sin(d*x + c)^3) - 15*log(sin(d*x +
c) + 1) + 15*log(sin(d*x + c) - 1)) + 6*a^2*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)) + 12*a*b^2*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)) + 4*(3*tan(
d*x + c)^2 + 1)*a^3/tan(d*x + c)^3)/d

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Fricas [A]  time = 1.86584, size = 618, normalized size = 3.01 \begin{align*} -\frac{8 \,{\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 6 \,{\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 36 \, a b^{2} \cos \left (d x + c\right ) - 12 \,{\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, b^{3} - 8 \,{\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left ({\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \,{\left ({\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(8*(a^3 + 12*a*b^2)*cos(d*x + c)^5 + 6*(6*a^2*b + 5*b^3)*cos(d*x + c)^4 + 36*a*b^2*cos(d*x + c) - 12*(a^
3 + 12*a*b^2)*cos(d*x + c)^3 + 6*b^3 - 8*(6*a^2*b + 5*b^3)*cos(d*x + c)^2 - 3*((6*a^2*b + 5*b^3)*cos(d*x + c)^
4 - (6*a^2*b + 5*b^3)*cos(d*x + c)^2)*log(sin(d*x + c) + 1)*sin(d*x + c) + 3*((6*a^2*b + 5*b^3)*cos(d*x + c)^4
- (6*a^2*b + 5*b^3)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)*sin(d*x + c))/((d*cos(d*x + c)^4 - d*cos(d*x + c)^
2)*sin(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)**4*(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.35067, size = 487, normalized size = 2.38 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 45 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 63 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 27 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \,{\left (6 \, a^{2} b + 5 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 12 \,{\left (6 \, a^{2} b + 5 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{24 \,{\left (6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac{9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 45 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 63 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 27 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1
/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c) - 45*a^2*b*tan(1/2*d*x + 1/2*c) + 63*a*b^2*tan(1/2*d*x + 1/2*c)
- 27*b^3*tan(1/2*d*x + 1/2*c) + 12*(6*a^2*b + 5*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*(6*a^2*b + 5*b^3
)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 24*(6*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b
^2*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - (9*a^3*tan(1/2*d*x + 1/2*
c)^2 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 63*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 27*b^3*tan(1/2*d*x + 1/2*c)^2 + a^3
+ 3*a^2*b + 3*a*b^2 + b^3)/tan(1/2*d*x + 1/2*c)^3)/d