3.72 \(\int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=259 \[ \frac{3 a^2 b \sec ^7(c+d x)}{7 d}+\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac{5 a^3 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{5 a^3 \tan (c+d x) \sec (c+d x)}{16 d}-\frac{15 a b^2 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{3 a b^2 \tan (c+d x) \sec ^7(c+d x)}{8 d}-\frac{a b^2 \tan (c+d x) \sec ^5(c+d x)}{16 d}-\frac{5 a b^2 \tan (c+d x) \sec ^3(c+d x)}{64 d}-\frac{15 a b^2 \tan (c+d x) \sec (c+d x)}{128 d}+\frac{b^3 \sec ^9(c+d x)}{9 d}-\frac{b^3 \sec ^7(c+d x)}{7 d} \]

[Out]

(5*a^3*ArcTanh[Sin[c + d*x]])/(16*d) - (15*a*b^2*ArcTanh[Sin[c + d*x]])/(128*d) + (3*a^2*b*Sec[c + d*x]^7)/(7*
d) - (b^3*Sec[c + d*x]^7)/(7*d) + (b^3*Sec[c + d*x]^9)/(9*d) + (5*a^3*Sec[c + d*x]*Tan[c + d*x])/(16*d) - (15*
a*b^2*Sec[c + d*x]*Tan[c + d*x])/(128*d) + (5*a^3*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) - (5*a*b^2*Sec[c + d*x]^
3*Tan[c + d*x])/(64*d) + (a^3*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) - (a*b^2*Sec[c + d*x]^5*Tan[c + d*x])/(16*d)
+ (3*a*b^2*Sec[c + d*x]^7*Tan[c + d*x])/(8*d)

________________________________________________________________________________________

Rubi [A]  time = 0.267958, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3090, 3768, 3770, 2606, 30, 2611, 14} \[ \frac{3 a^2 b \sec ^7(c+d x)}{7 d}+\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac{5 a^3 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{5 a^3 \tan (c+d x) \sec (c+d x)}{16 d}-\frac{15 a b^2 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{3 a b^2 \tan (c+d x) \sec ^7(c+d x)}{8 d}-\frac{a b^2 \tan (c+d x) \sec ^5(c+d x)}{16 d}-\frac{5 a b^2 \tan (c+d x) \sec ^3(c+d x)}{64 d}-\frac{15 a b^2 \tan (c+d x) \sec (c+d x)}{128 d}+\frac{b^3 \sec ^9(c+d x)}{9 d}-\frac{b^3 \sec ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(5*a^3*ArcTanh[Sin[c + d*x]])/(16*d) - (15*a*b^2*ArcTanh[Sin[c + d*x]])/(128*d) + (3*a^2*b*Sec[c + d*x]^7)/(7*
d) - (b^3*Sec[c + d*x]^7)/(7*d) + (b^3*Sec[c + d*x]^9)/(9*d) + (5*a^3*Sec[c + d*x]*Tan[c + d*x])/(16*d) - (15*
a*b^2*Sec[c + d*x]*Tan[c + d*x])/(128*d) + (5*a^3*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) - (5*a*b^2*Sec[c + d*x]^
3*Tan[c + d*x])/(64*d) + (a^3*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) - (a*b^2*Sec[c + d*x]^5*Tan[c + d*x])/(16*d)
+ (3*a*b^2*Sec[c + d*x]^7*Tan[c + d*x])/(8*d)

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && 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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=\int \left (a^3 \sec ^7(c+d x)+3 a^2 b \sec ^7(c+d x) \tan (c+d x)+3 a b^2 \sec ^7(c+d x) \tan ^2(c+d x)+b^3 \sec ^7(c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sec ^7(c+d x) \, dx+\left (3 a^2 b\right ) \int \sec ^7(c+d x) \tan (c+d x) \, dx+\left (3 a b^2\right ) \int \sec ^7(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sec ^7(c+d x) \tan ^3(c+d x) \, dx\\ &=\frac{a^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{3 a b^2 \sec ^7(c+d x) \tan (c+d x)}{8 d}+\frac{1}{6} \left (5 a^3\right ) \int \sec ^5(c+d x) \, dx-\frac{1}{8} \left (3 a b^2\right ) \int \sec ^7(c+d x) \, dx+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int x^6 \, dx,x,\sec (c+d x)\right )}{d}+\frac{b^3 \operatorname{Subst}\left (\int x^6 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{3 a^2 b \sec ^7(c+d x)}{7 d}+\frac{5 a^3 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{a^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{a b^2 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{3 a b^2 \sec ^7(c+d x) \tan (c+d x)}{8 d}+\frac{1}{8} \left (5 a^3\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{16} \left (5 a b^2\right ) \int \sec ^5(c+d x) \, dx+\frac{b^3 \operatorname{Subst}\left (\int \left (-x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{3 a^2 b \sec ^7(c+d x)}{7 d}-\frac{b^3 \sec ^7(c+d x)}{7 d}+\frac{b^3 \sec ^9(c+d x)}{9 d}+\frac{5 a^3 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{5 a^3 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac{5 a b^2 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{a^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{a b^2 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{3 a b^2 \sec ^7(c+d x) \tan (c+d x)}{8 d}+\frac{1}{16} \left (5 a^3\right ) \int \sec (c+d x) \, dx-\frac{1}{64} \left (15 a b^2\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{3 a^2 b \sec ^7(c+d x)}{7 d}-\frac{b^3 \sec ^7(c+d x)}{7 d}+\frac{b^3 \sec ^9(c+d x)}{9 d}+\frac{5 a^3 \sec (c+d x) \tan (c+d x)}{16 d}-\frac{15 a b^2 \sec (c+d x) \tan (c+d x)}{128 d}+\frac{5 a^3 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac{5 a b^2 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{a^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{a b^2 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{3 a b^2 \sec ^7(c+d x) \tan (c+d x)}{8 d}-\frac{1}{128} \left (15 a b^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{15 a b^2 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{3 a^2 b \sec ^7(c+d x)}{7 d}-\frac{b^3 \sec ^7(c+d x)}{7 d}+\frac{b^3 \sec ^9(c+d x)}{9 d}+\frac{5 a^3 \sec (c+d x) \tan (c+d x)}{16 d}-\frac{15 a b^2 \sec (c+d x) \tan (c+d x)}{128 d}+\frac{5 a^3 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac{5 a b^2 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{a^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{a b^2 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{3 a b^2 \sec ^7(c+d x) \tan (c+d x)}{8 d}\\ \end{align*}

Mathematica [B]  time = 4.10923, size = 810, normalized size = 3.13 \[ \frac{\sec ^9(c+d x) \left (-211680 \cos (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 ) a^3-90720 \cos (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 ) a^3-22680 \cos (7 (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^3-2520 \cos (9 (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^3+211680 \cos (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 ) a^3+90720 \cos (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 ) a^3+22680 \cos (7 (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^3+2520 \cos (9 (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^3+223776 \sin (2 (c+d x)) a^3+167328 \sin (4 (c+d x)) a^3+43680 \sin (6 (c+d x)) a^3+5040 \sin (8 (c+d x)) a^3+442368 b a^2+79380 b^2 \cos (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 ) a+34020 b^2 \cos (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 ) a+8505 b^2 \cos (7 (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+945 b^2 \cos (9 (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-39690 \left (8 a^2-3 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )\right ) a-79380 b^2 \cos (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 ) a-34020 b^2 \cos (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 ) a-8505 b^2 \cos (7 (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-945 b^2 \cos (9 (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+303156 b^2 \sin (2 (c+d x)) a-62748 b^2 \sin (4 (c+d x)) a-16380 b^2 \sin (6 (c+d x)) a-1890 b^2 \sin (8 (c+d x)) a+81920 b^3+147456 \left (3 a^2 b-b^3\right ) \cos (2 (c+d x))\right )}{2064384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^9*(442368*a^2*b + 81920*b^3 + 147456*(3*a^2*b - b^3)*Cos[2*(c + d*x)] - 211680*a^3*Cos[3*(c + d*
x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 79380*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]] - 90720*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 34020*a*b^2*Cos[5*(c + d*x)]*Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 22680*a^3*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
8505*a*b^2*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2520*a^3*Cos[9*(c + d*x)]*Log[Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]] + 945*a*b^2*Cos[9*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 39690*a*(
8*a^2 - 3*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]]) + 211680*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 79380*a*b^2*Cos[3*(c + d*x)]*Log[
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 90720*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 3
4020*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 22680*a^3*Cos[7*(c + d*x)]*Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]] - 8505*a*b^2*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 2520*a^3
*Cos[9*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 945*a*b^2*Cos[9*(c + d*x)]*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] + 223776*a^3*Sin[2*(c + d*x)] + 303156*a*b^2*Sin[2*(c + d*x)] + 167328*a^3*Sin[4*(c + d*x)]
- 62748*a*b^2*Sin[4*(c + d*x)] + 43680*a^3*Sin[6*(c + d*x)] - 16380*a*b^2*Sin[6*(c + d*x)] + 5040*a^3*Sin[8*(
c + d*x)] - 1890*a*b^2*Sin[8*(c + d*x)]))/(2064384*d)

________________________________________________________________________________________

Maple [A]  time = 0.139, size = 399, normalized size = 1.5 \begin{align*}{\frac{{a}^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{5}\tan \left ( dx+c \right ) }{6\,d}}+{\frac{5\,{a}^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{24\,d}}+{\frac{5\,{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{3\,{a}^{2}b}{7\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+{\frac{5\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{15\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{64\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{15\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{128\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{15\,a{b}^{2}\sin \left ( dx+c \right ) }{128\,d}}-{\frac{15\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{128\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{9\,d \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+{\frac{5\,{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{63\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{21\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{63\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{63\,d\cos \left ( dx+c \right ) }}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{63\,d}}-{\frac{2\,{b}^{3}\cos \left ( dx+c \right ) }{63\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/6*a^3*sec(d*x+c)^5*tan(d*x+c)/d+5/24*a^3*sec(d*x+c)^3*tan(d*x+c)/d+5/16*a^3*sec(d*x+c)*tan(d*x+c)/d+5/16/d*a
^3*ln(sec(d*x+c)+tan(d*x+c))+3/7/d*a^2*b/cos(d*x+c)^7+3/8/d*a*b^2*sin(d*x+c)^3/cos(d*x+c)^8+5/16/d*a*b^2*sin(d
*x+c)^3/cos(d*x+c)^6+15/64/d*a*b^2*sin(d*x+c)^3/cos(d*x+c)^4+15/128/d*a*b^2*sin(d*x+c)^3/cos(d*x+c)^2+15/128*a
*b^2*sin(d*x+c)/d-15/128/d*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/9/d*b^3*sin(d*x+c)^4/cos(d*x+c)^9+5/63/d*b^3*sin(
d*x+c)^4/cos(d*x+c)^7+1/21/d*b^3*sin(d*x+c)^4/cos(d*x+c)^5+1/63/d*b^3*sin(d*x+c)^4/cos(d*x+c)^3-1/63/d*b^3*sin
(d*x+c)^4/cos(d*x+c)-1/63/d*cos(d*x+c)*sin(d*x+c)^2*b^3-2/63*b^3*cos(d*x+c)/d

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Maxima [A]  time = 1.21562, size = 335, normalized size = 1.29 \begin{align*} \frac{63 \, a b^{2}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 168 \, a^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac{6912 \, a^{2} b}{\cos \left (d x + c\right )^{7}} - \frac{256 \,{\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} b^{3}}{\cos \left (d x + c\right )^{9}}}{16128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16128*(63*a*b^2*(2*(15*sin(d*x + c)^7 - 55*sin(d*x + c)^5 + 73*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x +
c)^8 - 4*sin(d*x + c)^6 + 6*sin(d*x + c)^4 - 4*sin(d*x + c)^2 + 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x
+ c) - 1)) - 168*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x
+ c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 6912*a^2*b/cos(d*x +
c)^7 - 256*(9*cos(d*x + c)^2 - 7)*b^3/cos(d*x + c)^9)/d

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Fricas [A]  time = 0.581469, size = 481, normalized size = 1.86 \begin{align*} \frac{315 \,{\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{9} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \,{\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{9} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 1792 \, b^{3} + 2304 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 42 \,{\left (15 \,{\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 10 \,{\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 144 \, a b^{2} \cos \left (d x + c\right ) + 8 \,{\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{16128 \, d \cos \left (d x + c\right )^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16128*(315*(8*a^3 - 3*a*b^2)*cos(d*x + c)^9*log(sin(d*x + c) + 1) - 315*(8*a^3 - 3*a*b^2)*cos(d*x + c)^9*log
(-sin(d*x + c) + 1) + 1792*b^3 + 2304*(3*a^2*b - b^3)*cos(d*x + c)^2 + 42*(15*(8*a^3 - 3*a*b^2)*cos(d*x + c)^7
+ 10*(8*a^3 - 3*a*b^2)*cos(d*x + c)^5 + 144*a*b^2*cos(d*x + c) + 8*(8*a^3 - 3*a*b^2)*cos(d*x + c)^3)*sin(d*x
+ c))/(d*cos(d*x + c)^9)

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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(sec(d*x+c)**10*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.21515, size = 806, normalized size = 3.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/8064*(315*(8*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 315*(8*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) + 2*(5544*a^3*tan(1/2*d*x + 1/2*c)^17 + 945*a*b^2*tan(1/2*d*x + 1/2*c)^17 - 24192*a^2*b*tan(1/2*
d*x + 1/2*c)^16 - 15792*a^3*tan(1/2*d*x + 1/2*c)^15 + 24066*a*b^2*tan(1/2*d*x + 1/2*c)^15 + 48384*a^2*b*tan(1/
2*d*x + 1/2*c)^14 - 16128*b^3*tan(1/2*d*x + 1/2*c)^14 + 29232*a^3*tan(1/2*d*x + 1/2*c)^13 + 31374*a*b^2*tan(1/
2*d*x + 1/2*c)^13 - 145152*a^2*b*tan(1/2*d*x + 1/2*c)^12 - 26880*b^3*tan(1/2*d*x + 1/2*c)^12 - 33264*a^3*tan(1
/2*d*x + 1/2*c)^11 + 54810*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 241920*a^2*b*tan(1/2*d*x + 1/2*c)^10 - 80640*b^3*ta
n(1/2*d*x + 1/2*c)^10 - 193536*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 48384*b^3*tan(1/2*d*x + 1/2*c)^8 + 33264*a^3*tan
(1/2*d*x + 1/2*c)^7 - 54810*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 145152*a^2*b*tan(1/2*d*x + 1/2*c)^6 - 48384*b^3*tan
(1/2*d*x + 1/2*c)^6 - 29232*a^3*tan(1/2*d*x + 1/2*c)^5 - 31374*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 76032*a^2*b*tan(
1/2*d*x + 1/2*c)^4 - 6912*b^3*tan(1/2*d*x + 1/2*c)^4 + 15792*a^3*tan(1/2*d*x + 1/2*c)^3 - 24066*a*b^2*tan(1/2*
d*x + 1/2*c)^3 + 6912*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 2304*b^3*tan(1/2*d*x + 1/2*c)^2 - 5544*a^3*tan(1/2*d*x +
1/2*c) - 945*a*b^2*tan(1/2*d*x + 1/2*c) - 3456*a^2*b + 256*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^9)/d