3.107 \(\int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=391 \[ \frac{10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac{2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a^3 b^2 \tan (c+d x) \sec ^5(c+d x)}{3 d}-\frac{5 a^3 b^2 \tan (c+d x) \sec ^3(c+d x)}{12 d}-\frac{5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^4 b \sec ^5(c+d x)}{d}+\frac{3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^5 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a^5 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{15 a b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{5 a b^4 \tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac{5 a b^4 \tan (c+d x) \sec ^5(c+d x)}{16 d}+\frac{5 a b^4 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac{15 a b^4 \tan (c+d x) \sec (c+d x)}{128 d}+\frac{b^5 \sec ^9(c+d x)}{9 d}-\frac{2 b^5 \sec ^7(c+d x)}{7 d}+\frac{b^5 \sec ^5(c+d x)}{5 d} \]

[Out]

(3*a^5*ArcTanh[Sin[c + d*x]])/(8*d) - (5*a^3*b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (15*a*b^4*ArcTanh[Sin[c + d*x]
])/(128*d) + (a^4*b*Sec[c + d*x]^5)/d - (2*a^2*b^3*Sec[c + d*x]^5)/d + (b^5*Sec[c + d*x]^5)/(5*d) + (10*a^2*b^
3*Sec[c + d*x]^7)/(7*d) - (2*b^5*Sec[c + d*x]^7)/(7*d) + (b^5*Sec[c + d*x]^9)/(9*d) + (3*a^5*Sec[c + d*x]*Tan[
c + d*x])/(8*d) - (5*a^3*b^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (15*a*b^4*Sec[c + d*x]*Tan[c + d*x])/(128*d) +
(a^5*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) - (5*a^3*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(12*d) + (5*a*b^4*Sec[c + d
*x]^3*Tan[c + d*x])/(64*d) + (5*a^3*b^2*Sec[c + d*x]^5*Tan[c + d*x])/(3*d) - (5*a*b^4*Sec[c + d*x]^5*Tan[c + d
*x])/(16*d) + (5*a*b^4*Sec[c + d*x]^5*Tan[c + d*x]^3)/(8*d)

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Rubi [A]  time = 0.389182, antiderivative size = 391, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3090, 3768, 3770, 2606, 30, 2611, 14, 270} \[ \frac{10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac{2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a^3 b^2 \tan (c+d x) \sec ^5(c+d x)}{3 d}-\frac{5 a^3 b^2 \tan (c+d x) \sec ^3(c+d x)}{12 d}-\frac{5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^4 b \sec ^5(c+d x)}{d}+\frac{3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^5 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a^5 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{15 a b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{5 a b^4 \tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac{5 a b^4 \tan (c+d x) \sec ^5(c+d x)}{16 d}+\frac{5 a b^4 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac{15 a b^4 \tan (c+d x) \sec (c+d x)}{128 d}+\frac{b^5 \sec ^9(c+d x)}{9 d}-\frac{2 b^5 \sec ^7(c+d x)}{7 d}+\frac{b^5 \sec ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(3*a^5*ArcTanh[Sin[c + d*x]])/(8*d) - (5*a^3*b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (15*a*b^4*ArcTanh[Sin[c + d*x]
])/(128*d) + (a^4*b*Sec[c + d*x]^5)/d - (2*a^2*b^3*Sec[c + d*x]^5)/d + (b^5*Sec[c + d*x]^5)/(5*d) + (10*a^2*b^
3*Sec[c + d*x]^7)/(7*d) - (2*b^5*Sec[c + d*x]^7)/(7*d) + (b^5*Sec[c + d*x]^9)/(9*d) + (3*a^5*Sec[c + d*x]*Tan[
c + d*x])/(8*d) - (5*a^3*b^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (15*a*b^4*Sec[c + d*x]*Tan[c + d*x])/(128*d) +
(a^5*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) - (5*a^3*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(12*d) + (5*a*b^4*Sec[c + d
*x]^3*Tan[c + d*x])/(64*d) + (5*a^3*b^2*Sec[c + d*x]^5*Tan[c + d*x])/(3*d) - (5*a*b^4*Sec[c + d*x]^5*Tan[c + d
*x])/(16*d) + (5*a*b^4*Sec[c + d*x]^5*Tan[c + d*x]^3)/(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]]

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]

Rubi steps

\begin{align*} \int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \sec ^5(c+d x)+5 a^4 b \sec ^5(c+d x) \tan (c+d x)+10 a^3 b^2 \sec ^5(c+d x) \tan ^2(c+d x)+10 a^2 b^3 \sec ^5(c+d x) \tan ^3(c+d x)+5 a b^4 \sec ^5(c+d x) \tan ^4(c+d x)+b^5 \sec ^5(c+d x) \tan ^5(c+d x)\right ) \, dx\\ &=a^5 \int \sec ^5(c+d x) \, dx+\left (5 a^4 b\right ) \int \sec ^5(c+d x) \tan (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sec ^5(c+d x) \tan ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sec ^5(c+d x) \tan ^4(c+d x) \, dx+b^5 \int \sec ^5(c+d x) \tan ^5(c+d x) \, dx\\ &=\frac{a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}+\frac{5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac{1}{4} \left (3 a^5\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{3} \left (5 a^3 b^2\right ) \int \sec ^5(c+d x) \, dx-\frac{1}{8} \left (15 a b^4\right ) \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx+\frac{\left (5 a^4 b\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (10 a^2 b^3\right ) \operatorname{Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac{b^5 \operatorname{Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a^4 b \sec ^5(c+d x)}{d}+\frac{3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac{5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac{5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac{1}{8} \left (3 a^5\right ) \int \sec (c+d x) \, dx-\frac{1}{4} \left (5 a^3 b^2\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{16} \left (5 a b^4\right ) \int \sec ^5(c+d x) \, dx+\frac{\left (10 a^2 b^3\right ) \operatorname{Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac{b^5 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 b \sec ^5(c+d x)}{d}-\frac{2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac{b^5 \sec ^5(c+d x)}{5 d}+\frac{10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac{2 b^5 \sec ^7(c+d x)}{7 d}+\frac{b^5 \sec ^9(c+d x)}{9 d}+\frac{3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac{5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac{5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}-\frac{1}{8} \left (5 a^3 b^2\right ) \int \sec (c+d x) \, dx+\frac{1}{64} \left (15 a b^4\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 b \sec ^5(c+d x)}{d}-\frac{2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac{b^5 \sec ^5(c+d x)}{5 d}+\frac{10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac{2 b^5 \sec ^7(c+d x)}{7 d}+\frac{b^5 \sec ^9(c+d x)}{9 d}+\frac{3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{15 a b^4 \sec (c+d x) \tan (c+d x)}{128 d}+\frac{a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac{5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac{5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac{1}{128} \left (15 a b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{15 a b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{a^4 b \sec ^5(c+d x)}{d}-\frac{2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac{b^5 \sec ^5(c+d x)}{5 d}+\frac{10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac{2 b^5 \sec ^7(c+d x)}{7 d}+\frac{b^5 \sec ^9(c+d x)}{9 d}+\frac{3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{15 a b^4 \sec (c+d x) \tan (c+d x)}{128 d}+\frac{a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac{5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac{5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}\\ \end{align*}

Mathematica [A]  time = 2.21531, size = 331, normalized size = 0.85 \[ \frac{1260 a \left (2320 a^2 b^2+656 a^4+845 b^4\right ) \tan (c+d x) \sec ^7(c+d x)-40320 a \left (-80 a^2 b^2+48 a^4+15 b^4\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec ^9(c+d x) \left (453600 a^3 b^2 \sin (4 (c+d x))-218400 a^3 b^2 \sin (6 (c+d x))-25200 a^3 b^2 \sin (8 (c+d x))+73728 \left (-20 a^2 b^3+35 a^4 b-3 b^5\right ) \cos (2 (c+d x))+129024 \left (-10 a^2 b^3+5 a^4 b+b^5\right ) \cos (4 (c+d x))-184320 a^2 b^3+1935360 a^4 b+372960 a^5 \sin (4 (c+d x))+131040 a^5 \sin (6 (c+d x))+15120 a^5 \sin (8 (c+d x))-488250 a b^4 \sin (4 (c+d x))+40950 a b^4 \sin (6 (c+d x))+4725 a b^4 \sin (8 (c+d x))+223232 b^5\right )}{5160960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(-40320*a*(48*a^4 - 80*a^2*b^2 + 15*b^4)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]]) + Sec[c + d*x]^9*(1935360*a^4*b - 184320*a^2*b^3 + 223232*b^5 + 73728*(35*a^4*b - 20*a^2*b^3
- 3*b^5)*Cos[2*(c + d*x)] + 129024*(5*a^4*b - 10*a^2*b^3 + b^5)*Cos[4*(c + d*x)] + 372960*a^5*Sin[4*(c + d*x)]
+ 453600*a^3*b^2*Sin[4*(c + d*x)] - 488250*a*b^4*Sin[4*(c + d*x)] + 131040*a^5*Sin[6*(c + d*x)] - 218400*a^3*
b^2*Sin[6*(c + d*x)] + 40950*a*b^4*Sin[6*(c + d*x)] + 15120*a^5*Sin[8*(c + d*x)] - 25200*a^3*b^2*Sin[8*(c + d*
x)] + 4725*a*b^4*Sin[8*(c + d*x)]) + 1260*a*(656*a^4 + 2320*a^2*b^2 + 845*b^4)*Sec[c + d*x]^7*Tan[c + d*x])/(5
160960*d)

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Maple [A]  time = 0.266, size = 688, normalized size = 1.8 \begin{align*} \text{result too large to display} \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))^5,x)

[Out]

-4/7*a^2*b^3*cos(d*x+c)/d-15/128*a*b^4*sin(d*x+c)/d-5/128*a*b^4*sin(d*x+c)^3/d+8/315*b^5*cos(d*x+c)/d+5/8/d*a*
b^4*sin(d*x+c)^5/cos(d*x+c)^8+10/7/d*a^2*b^3*sin(d*x+c)^4/cos(d*x+c)^7+6/7/d*a^2*b^3*sin(d*x+c)^4/cos(d*x+c)^5
+5/8*a^3*b^2*sin(d*x+c)/d+3/8*a^5*sec(d*x+c)*tan(d*x+c)/d+1/4*a^5*sec(d*x+c)^3*tan(d*x+c)/d-5/8/d*a^3*b^2*ln(s
ec(d*x+c)+tan(d*x+c))-1/315/d*b^5*sin(d*x+c)^6/cos(d*x+c)^3+1/21/d*b^5*sin(d*x+c)^6/cos(d*x+c)^7-2/7/d*cos(d*x
+c)*sin(d*x+c)^2*a^2*b^3+5/3/d*a^3*b^2*sin(d*x+c)^3/cos(d*x+c)^6+5/16/d*a*b^4*sin(d*x+c)^5/cos(d*x+c)^6+5/4/d*
a^3*b^2*sin(d*x+c)^3/cos(d*x+c)^4+3/8/d*a^5*ln(sec(d*x+c)+tan(d*x+c))+1/9/d*b^5*sin(d*x+c)^6/cos(d*x+c)^9+1/d*
a^4*b/cos(d*x+c)^5+15/128/d*a*b^4*ln(sec(d*x+c)+tan(d*x+c))+1/105/d*b^5*sin(d*x+c)^6/cos(d*x+c)+1/105/d*b^5*co
s(d*x+c)*sin(d*x+c)^4+4/315/d*cos(d*x+c)*sin(d*x+c)^2*b^5+1/105/d*b^5*sin(d*x+c)^6/cos(d*x+c)^5-2/7/d*a^2*b^3*
sin(d*x+c)^4/cos(d*x+c)+2/7/d*a^2*b^3*sin(d*x+c)^4/cos(d*x+c)^3+5/64/d*a*b^4*sin(d*x+c)^5/cos(d*x+c)^4+5/8/d*a
^3*b^2*sin(d*x+c)^3/cos(d*x+c)^2-5/128/d*a*b^4*sin(d*x+c)^5/cos(d*x+c)^2

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Maxima [A]  time = 1.22289, size = 486, normalized size = 1.24 \begin{align*} -\frac{1575 \, a b^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{7} - 11 \, \sin \left (d x + c\right )^{5} - 11 \, \sin \left (d x + c\right )^{3} + 3 \, \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} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8400 \, a^{3} b^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \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} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 5040 \, a^{5}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac{80640 \, a^{4} b}{\cos \left (d x + c\right )^{5}} + \frac{23040 \,{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} a^{2} b^{3}}{\cos \left (d x + c\right )^{7}} - \frac{256 \,{\left (63 \, \cos \left (d x + c\right )^{4} - 90 \, \cos \left (d x + c\right )^{2} + 35\right )} b^{5}}{\cos \left (d x + c\right )^{9}}}{80640 \, 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))^5,x, algorithm="maxima")

[Out]

-1/80640*(1575*a*b^4*(2*(3*sin(d*x + c)^7 - 11*sin(d*x + c)^5 - 11*sin(d*x + c)^3 + 3*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) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x
+ c) - 1)) - 8400*a^3*b^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*
x + c)^4 + 3*sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 5040*a^5*(2*(3*sin(d*x
+ c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 80640*a^4*b/cos(d*x + c)^5 + 23040*(7*cos(d*x + c)^2 - 5)*a^2*b^3/cos(d*x + c)^7 - 256*(63*cos(d*x
+ c)^4 - 90*cos(d*x + c)^2 + 35)*b^5/cos(d*x + c)^9)/d

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Fricas [A]  time = 0.628745, size = 643, normalized size = 1.64 \begin{align*} \frac{315 \,{\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{9} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \,{\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{9} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8960 \, b^{5} + 16128 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 23040 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 210 \,{\left (3 \,{\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} + 240 \, a b^{4} \cos \left (d x + c\right ) + 2 \,{\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 40 \,{\left (16 \, a^{3} b^{2} - 9 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{80640 \, 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))^5,x, algorithm="fricas")

[Out]

1/80640*(315*(48*a^5 - 80*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^9*log(sin(d*x + c) + 1) - 315*(48*a^5 - 80*a^3*b^2
+ 15*a*b^4)*cos(d*x + c)^9*log(-sin(d*x + c) + 1) + 8960*b^5 + 16128*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)
^4 + 23040*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 210*(3*(48*a^5 - 80*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^7 + 240*a*b
^4*cos(d*x + c) + 2*(48*a^5 - 80*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5 + 40*(16*a^3*b^2 - 9*a*b^4)*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))**5,x)

[Out]

Timed out

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Giac [B]  time = 1.33802, size = 1199, normalized size = 3.07 \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))^5,x, algorithm="giac")

[Out]

1/40320*(315*(48*a^5 - 80*a^3*b^2 + 15*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 315*(48*a^5 - 80*a^3*b^2 +
15*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(25200*a^5*tan(1/2*d*x + 1/2*c)^17 + 25200*a^3*b^2*tan(1/2*d*
x + 1/2*c)^17 - 4725*a*b^4*tan(1/2*d*x + 1/2*c)^17 - 201600*a^4*b*tan(1/2*d*x + 1/2*c)^16 - 110880*a^5*tan(1/2
*d*x + 1/2*c)^15 + 319200*a^3*b^2*tan(1/2*d*x + 1/2*c)^15 + 40950*a*b^4*tan(1/2*d*x + 1/2*c)^15 + 806400*a^4*b
*tan(1/2*d*x + 1/2*c)^14 - 806400*a^2*b^3*tan(1/2*d*x + 1/2*c)^14 + 191520*a^5*tan(1/2*d*x + 1/2*c)^13 - 45360
0*a^3*b^2*tan(1/2*d*x + 1/2*c)^13 + 488250*a*b^4*tan(1/2*d*x + 1/2*c)^13 - 1612800*a^4*b*tan(1/2*d*x + 1/2*c)^
12 + 806400*a^2*b^3*tan(1/2*d*x + 1/2*c)^12 - 215040*b^5*tan(1/2*d*x + 1/2*c)^12 - 151200*a^5*tan(1/2*d*x + 1/
2*c)^11 - 151200*a^3*b^2*tan(1/2*d*x + 1/2*c)^11 + 532350*a*b^4*tan(1/2*d*x + 1/2*c)^11 + 2419200*a^4*b*tan(1/
2*d*x + 1/2*c)^10 - 806400*a^2*b^3*tan(1/2*d*x + 1/2*c)^10 - 322560*b^5*tan(1/2*d*x + 1/2*c)^10 - 2661120*a^4*
b*tan(1/2*d*x + 1/2*c)^8 + 2096640*a^2*b^3*tan(1/2*d*x + 1/2*c)^8 - 451584*b^5*tan(1/2*d*x + 1/2*c)^8 + 151200
*a^5*tan(1/2*d*x + 1/2*c)^7 + 151200*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 - 532350*a*b^4*tan(1/2*d*x + 1/2*c)^7 + 17
74080*a^4*b*tan(1/2*d*x + 1/2*c)^6 - 1128960*a^2*b^3*tan(1/2*d*x + 1/2*c)^6 - 129024*b^5*tan(1/2*d*x + 1/2*c)^
6 - 191520*a^5*tan(1/2*d*x + 1/2*c)^5 + 453600*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 488250*a*b^4*tan(1/2*d*x + 1/2
*c)^5 - 645120*a^4*b*tan(1/2*d*x + 1/2*c)^4 + 23040*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 36864*b^5*tan(1/2*d*x + 1
/2*c)^4 + 110880*a^5*tan(1/2*d*x + 1/2*c)^3 - 319200*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 40950*a*b^4*tan(1/2*d*x
+ 1/2*c)^3 + 161280*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 207360*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 9216*b^5*tan(1/2*d*
x + 1/2*c)^2 - 25200*a^5*tan(1/2*d*x + 1/2*c) - 25200*a^3*b^2*tan(1/2*d*x + 1/2*c) + 4725*a*b^4*tan(1/2*d*x +
1/2*c) - 40320*a^4*b + 23040*a^2*b^3 - 1024*b^5)/(tan(1/2*d*x + 1/2*c)^2 - 1)^9)/d