3.195 \(\int \csc ^6(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=279 \[ -\frac{3 a^2 b \csc ^5(c+d x)}{5 d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{3 a^2 b \csc (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{2 a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{3 a b^2 \cot ^5(c+d x)}{5 d}-\frac{3 a b^2 \cot ^3(c+d x)}{d}-\frac{9 a b^2 \cot (c+d x)}{d}-\frac{7 b^3 \csc ^5(c+d x)}{10 d}-\frac{7 b^3 \csc ^3(c+d x)}{6 d}-\frac{7 b^3 \csc (c+d x)}{2 d}+\frac{7 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \csc ^5(c+d x) \sec ^2(c+d x)}{2 d} \]

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (7*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (9*a*b^2*Cot[
c + d*x])/d - (2*a^3*Cot[c + d*x]^3)/(3*d) - (3*a*b^2*Cot[c + d*x]^3)/d - (a^3*Cot[c + d*x]^5)/(5*d) - (3*a*b^
2*Cot[c + d*x]^5)/(5*d) - (3*a^2*b*Csc[c + d*x])/d - (7*b^3*Csc[c + d*x])/(2*d) - (a^2*b*Csc[c + d*x]^3)/d - (
7*b^3*Csc[c + d*x]^3)/(6*d) - (3*a^2*b*Csc[c + d*x]^5)/(5*d) - (7*b^3*Csc[c + d*x]^5)/(10*d) + (b^3*Csc[c + d*
x]^5*Sec[c + d*x]^2)/(2*d) + (3*a*b^2*Tan[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.316217, antiderivative size = 279, 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} \[ -\frac{3 a^2 b \csc ^5(c+d x)}{5 d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{3 a^2 b \csc (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{2 a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{3 a b^2 \cot ^5(c+d x)}{5 d}-\frac{3 a b^2 \cot ^3(c+d x)}{d}-\frac{9 a b^2 \cot (c+d x)}{d}-\frac{7 b^3 \csc ^5(c+d x)}{10 d}-\frac{7 b^3 \csc ^3(c+d x)}{6 d}-\frac{7 b^3 \csc (c+d x)}{2 d}+\frac{7 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \csc ^5(c+d x) \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + b*Sec[c + d*x])^3,x]

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (7*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (9*a*b^2*Cot[
c + d*x])/d - (2*a^3*Cot[c + d*x]^3)/(3*d) - (3*a*b^2*Cot[c + d*x]^3)/d - (a^3*Cot[c + d*x]^5)/(5*d) - (3*a*b^
2*Cot[c + d*x]^5)/(5*d) - (3*a^2*b*Csc[c + d*x])/d - (7*b^3*Csc[c + d*x])/(2*d) - (a^2*b*Csc[c + d*x]^3)/d - (
7*b^3*Csc[c + d*x]^3)/(6*d) - (3*a^2*b*Csc[c + d*x]^5)/(5*d) - (7*b^3*Csc[c + d*x]^5)/(10*d) + (b^3*Csc[c + d*
x]^5*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 ^6(c+d x) (a+b \sec (c+d x))^3 \, dx &=-\int (-b-a \cos (c+d x))^3 \csc ^6(c+d x) \sec ^3(c+d x) \, dx\\ &=\int \left (a^3 \csc ^6(c+d x)+3 a^2 b \csc ^6(c+d x) \sec (c+d x)+3 a b^2 \csc ^6(c+d x) \sec ^2(c+d x)+b^3 \csc ^6(c+d x) \sec ^3(c+d x)\right ) \, dx\\ &=a^3 \int \csc ^6(c+d x) \, dx+\left (3 a^2 b\right ) \int \csc ^6(c+d x) \sec (c+d x) \, dx+\left (3 a b^2\right ) \int \csc ^6(c+d x) \sec ^2(c+d x) \, dx+b^3 \int \csc ^6(c+d x) \sec ^3(c+d x) \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^6}{-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 )^3}{x^6} \, dx,x,\tan (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{x^8}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{a^3 \cot (c+d x)}{d}-\frac{2 a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}+\frac{b^3 \csc ^5(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+x^4+\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^6}+\frac{3}{x^4}+\frac{3}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (7 b^3\right ) \operatorname{Subst}\left (\int \frac{x^6}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=-\frac{a^3 \cot (c+d x)}{d}-\frac{9 a b^2 \cot (c+d x)}{d}-\frac{2 a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a b^2 \cot ^3(c+d x)}{d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{3 a b^2 \cot ^5(c+d x)}{5 d}-\frac{3 a^2 b \csc (c+d x)}{d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{3 a^2 b \csc ^5(c+d x)}{5 d}+\frac{b^3 \csc ^5(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 (7 b^3\right ) \operatorname{Subst}\left (\int \left (1+x^2+x^4+\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{9 a b^2 \cot (c+d x)}{d}-\frac{2 a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a b^2 \cot ^3(c+d x)}{d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{3 a b^2 \cot ^5(c+d x)}{5 d}-\frac{3 a^2 b \csc (c+d x)}{d}-\frac{7 b^3 \csc (c+d x)}{2 d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{7 b^3 \csc ^3(c+d x)}{6 d}-\frac{3 a^2 b \csc ^5(c+d x)}{5 d}-\frac{7 b^3 \csc ^5(c+d x)}{10 d}+\frac{b^3 \csc ^5(c+d x) \sec ^2(c+d x)}{2 d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{\left (7 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{7 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \cot (c+d x)}{d}-\frac{9 a b^2 \cot (c+d x)}{d}-\frac{2 a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a b^2 \cot ^3(c+d x)}{d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{3 a b^2 \cot ^5(c+d x)}{5 d}-\frac{3 a^2 b \csc (c+d x)}{d}-\frac{7 b^3 \csc (c+d x)}{2 d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{7 b^3 \csc ^3(c+d x)}{6 d}-\frac{3 a^2 b \csc ^5(c+d x)}{5 d}-\frac{7 b^3 \csc ^5(c+d x)}{10 d}+\frac{b^3 \csc ^5(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 = 1.41209, size = 812, normalized size = 2.91 \[ -\frac{\csc ^9\left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (16 \cos (3 (c+d x)) a^3-48 \cos (5 (c+d x)) a^3+16 \cos (7 (c+d x)) a^3+1176 b a^2-600 b \cos (4 (c+d x)) a^2+180 b \cos (6 (c+d x)) a^2+450 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (c+d x) a^2-450 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (c+d x) a^2+90 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (3 (c+d x)) a^2-90 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (3 (c+d x)) a^2-270 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (5 (c+d x)) a^2+270 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (5 (c+d x)) a^2+90 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (7 (c+d x)) a^2-90 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (7 (c+d x)) a^2+80 \left (5 a^2+18 b^2\right ) \cos (c+d x) a+288 b^2 \cos (3 (c+d x)) a-864 b^2 \cos (5 (c+d x)) a+288 b^2 \cos (7 (c+d x)) a+412 b^3+66 \left (7 b^3+6 a^2 b\right ) \cos (2 (c+d x))-700 b^3 \cos (4 (c+d x))+210 b^3 \cos (6 (c+d x))+525 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (c+d x)-525 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (c+d x)+105 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-105 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-315 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+315 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+105 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (7 (c+d x))-105 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sin (7 (c+d x))\right )}{61440 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]^6*(a + b*Sec[c + d*x])^3,x]

[Out]

-(Csc[(c + d*x)/2]^9*Sec[(c + d*x)/2]^5*(1176*a^2*b + 412*b^3 + 80*a*(5*a^2 + 18*b^2)*Cos[c + d*x] + 66*(6*a^2
*b + 7*b^3)*Cos[2*(c + d*x)] + 16*a^3*Cos[3*(c + d*x)] + 288*a*b^2*Cos[3*(c + d*x)] - 600*a^2*b*Cos[4*(c + d*x
)] - 700*b^3*Cos[4*(c + d*x)] - 48*a^3*Cos[5*(c + d*x)] - 864*a*b^2*Cos[5*(c + d*x)] + 180*a^2*b*Cos[6*(c + d*
x)] + 210*b^3*Cos[6*(c + d*x)] + 16*a^3*Cos[7*(c + d*x)] + 288*a*b^2*Cos[7*(c + d*x)] + 450*a^2*b*Log[Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 525*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 45
0*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 525*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]]*Sin[c + d*x] + 90*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 105*b^3*Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 90*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)
] - 105*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 270*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(
c + d*x)/2]]*Sin[5*(c + d*x)] - 315*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 270*a^2*b*
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 315*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*S
in[5*(c + d*x)] + 90*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[7*(c + d*x)] + 105*b^3*Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]]*Sin[7*(c + d*x)] - 90*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[7*(c + d*x)
] - 105*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[7*(c + d*x)]))/(61440*d*(-1 + Cot[(c + d*x)/2]^2)^2)

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Maple [A]  time = 0.051, size = 334, normalized size = 1.2 \begin{align*} -{\frac{8\,{a}^{3}\cot \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{3}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{3}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{3\,{a}^{2}b}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\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{3\,a{b}^{2}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) }}-{\frac{6\,a{b}^{2}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{24\,a{b}^{2}}{5\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{48\,a{b}^{2}\cot \left ( dx+c \right ) }{5\,d}}-{\frac{{b}^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,{b}^{3}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{7\,{b}^{3}}{6\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,{b}^{3}}{2\,d\sin \left ( dx+c \right ) }}+{\frac{7\,{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)^6*(a+b*sec(d*x+c))^3,x)

[Out]

-8/15*a^3*cot(d*x+c)/d-1/5/d*a^3*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^3*cot(d*x+c)*csc(d*x+c)^2-3/5/d*a^2*b/sin(d*
x+c)^5-1/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))-3/5/d*a*b^2/sin(d*x+c)^
5/cos(d*x+c)-6/5/d*a*b^2/sin(d*x+c)^3/cos(d*x+c)+24/5/d*a*b^2/sin(d*x+c)/cos(d*x+c)-48/5*a*b^2*cot(d*x+c)/d-1/
5/d*b^3/sin(d*x+c)^5/cos(d*x+c)^2-7/15/d*b^3/sin(d*x+c)^3/cos(d*x+c)^2+7/6/d*b^3/sin(d*x+c)/cos(d*x+c)^2-7/2/d
*b^3/sin(d*x+c)+7/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.997119, size = 311, normalized size = 1.11 \begin{align*} -\frac{b^{3}{\left (\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{6} - 70 \, \sin \left (d x + c\right )^{4} - 14 \, \sin \left (d x + c\right )^{2} - 6\right )}}{\sin \left (d x + c\right )^{7} - \sin \left (d x + c\right )^{5}} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} b{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} + 3\right )}}{\sin \left (d x + c\right )^{5}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a b^{2}{\left (\frac{15 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{5}} - 5 \, \tan \left (d x + c\right )\right )} + \frac{4 \,{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(b^3*(2*(105*sin(d*x + c)^6 - 70*sin(d*x + c)^4 - 14*sin(d*x + c)^2 - 6)/(sin(d*x + c)^7 - sin(d*x + c)^
5) - 105*log(sin(d*x + c) + 1) + 105*log(sin(d*x + c) - 1)) + 6*a^2*b*(2*(15*sin(d*x + c)^4 + 5*sin(d*x + c)^2
+ 3)/sin(d*x + c)^5 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 36*a*b^2*((15*tan(d*x + c)^4 + 5
*tan(d*x + c)^2 + 1)/tan(d*x + c)^5 - 5*tan(d*x + c)) + 4*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a^3/tan(
d*x + c)^5)/d

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Fricas [A]  time = 1.8618, size = 857, normalized size = 3.07 \begin{align*} -\frac{32 \,{\left (a^{3} + 18 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 30 \,{\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 80 \,{\left (a^{3} + 18 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 70 \,{\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 180 \, a b^{2} \cos \left (d x + c\right ) + 60 \,{\left (a^{3} + 18 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 30 \, b^{3} + 46 \,{\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 15 \,{\left ({\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \,{\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{4} +{\left (6 \, a^{2} b + 7 \, 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 ) + 15 \,{\left ({\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \,{\left (6 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{4} +{\left (6 \, a^{2} b + 7 \, 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 )}{60 \,{\left (d \cos \left (d x + c\right )^{6} - 2 \, 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)^6*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(32*(a^3 + 18*a*b^2)*cos(d*x + c)^7 + 30*(6*a^2*b + 7*b^3)*cos(d*x + c)^6 - 80*(a^3 + 18*a*b^2)*cos(d*x
+ c)^5 - 70*(6*a^2*b + 7*b^3)*cos(d*x + c)^4 - 180*a*b^2*cos(d*x + c) + 60*(a^3 + 18*a*b^2)*cos(d*x + c)^3 - 3
0*b^3 + 46*(6*a^2*b + 7*b^3)*cos(d*x + c)^2 - 15*((6*a^2*b + 7*b^3)*cos(d*x + c)^6 - 2*(6*a^2*b + 7*b^3)*cos(d
*x + c)^4 + (6*a^2*b + 7*b^3)*cos(d*x + c)^2)*log(sin(d*x + c) + 1)*sin(d*x + c) + 15*((6*a^2*b + 7*b^3)*cos(d
*x + c)^6 - 2*(6*a^2*b + 7*b^3)*cos(d*x + c)^4 + (6*a^2*b + 7*b^3)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)*sin(
d*x + c))/((d*cos(d*x + c)^6 - 2*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)**6*(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33031, size = 672, normalized size = 2.41 \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)^6*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/480*(3*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*b^3*
tan(1/2*d*x + 1/2*c)^5 + 25*a^3*tan(1/2*d*x + 1/2*c)^3 - 105*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 135*a*b^2*tan(1/2*
d*x + 1/2*c)^3 - 55*b^3*tan(1/2*d*x + 1/2*c)^3 + 150*a^3*tan(1/2*d*x + 1/2*c) - 990*a^2*b*tan(1/2*d*x + 1/2*c)
+ 1710*a*b^2*tan(1/2*d*x + 1/2*c) - 870*b^3*tan(1/2*d*x + 1/2*c) + 240*(6*a^2*b + 7*b^3)*log(abs(tan(1/2*d*x
+ 1/2*c) + 1)) - 240*(6*a^2*b + 7*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 480*(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 - (150*a^3*tan(1/2*d*x + 1/2*c)^4 + 990*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 1710*a*b^2*tan(1/2*d*x + 1/2
*c)^4 + 870*b^3*tan(1/2*d*x + 1/2*c)^4 + 25*a^3*tan(1/2*d*x + 1/2*c)^2 + 105*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 13
5*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 55*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^3 + 9*a^2*b + 9*a*b^2 + 3*b^3)/tan(1/2*d*
x + 1/2*c)^5)/d