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

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

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 30

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

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 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 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]

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*}

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Mathematica [B]  time = 4.02, 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)

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fricas [A]  time = 0.76, size = 192, normalized size = 0.74 \[ \frac {315 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{9} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{9} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 1792 \, b^{3} + 2304 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 42 \, {\left (15 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 10 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 144 \, a b^{2} \cos \left (d x + c\right ) + 8 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{16128 \, d \cos \left (d x + c\right )^{9}} \]

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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giac [B]  time = 0.91, size = 597, normalized size = 2.31 \[ \frac {315 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 315 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5544 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} + 945 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} - 24192 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{16} - 15792 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 24066 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 48384 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} - 16128 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} + 29232 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 31374 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 145152 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 26880 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 33264 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 54810 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 241920 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 80640 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 193536 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 48384 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 33264 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 54810 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 145152 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 48384 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 29232 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 31374 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 76032 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6912 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15792 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24066 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6912 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2304 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 5544 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 945 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3456 \, a^{2} b + 256 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{9}}}{8064 \, d} \]

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

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maple [A]  time = 21.07, size = 399, normalized size = 1.54 \[ \frac {a^{3} \left (\sec ^{5}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{6 d}+\frac {5 a^{3} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{24 d}+\frac {5 a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {3 a^{2} b}{7 d \cos \left (d x +c \right )^{7}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {5 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{6}}+\frac {15 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}+\frac {15 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}+\frac {15 a \,b^{2} \sin \left (d x +c \right )}{128 d}-\frac {15 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128 d}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )^{9}}+\frac {5 b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{7}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{21 d \cos \left (d x +c \right )^{5}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{3}}-\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )}-\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{63 d}-\frac {2 b^{3} \cos \left (d x +c \right )}{63 d} \]

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*b^2*a*sin(d*x+c)^3/cos(d*x+c)^8+5/16/d*b^2*a*sin(d
*x+c)^3/cos(d*x+c)^6+15/64/d*b^2*a*sin(d*x+c)^3/cos(d*x+c)^4+15/128/d*b^2*a*sin(d*x+c)^3/cos(d*x+c)^2+15/128*a
*b^2*sin(d*x+c)/d-15/128/d*b^2*a*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*b^3*cos(d*x+c)*sin(d*x+c)^2-2/63*b^3*cos(d*x+c)/d

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maxima [A]  time = 0.34, size = 248, normalized size = 0.96 \[ \frac {63 \, a b^{2} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 168 \, a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {6912 \, a^{2} b}{\cos \left (d x + c\right )^{7}} - \frac {256 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} b^{3}}{\cos \left (d x + c\right )^{9}}}{16128 \, d} \]

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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mupad [B]  time = 4.56, size = 547, normalized size = 2.11 \[ -\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {15\,a\,b^2}{64}-\frac {5\,a^3}{8}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {11\,a^3}{8}+\frac {15\,a\,b^2}{64}\right )+\frac {6\,a^2\,b}{7}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}\,\left (\frac {11\,a^3}{8}+\frac {15\,a\,b^2}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {191\,a\,b^2}{32}-\frac {47\,a^3}{12}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\left (\frac {191\,a\,b^2}{32}-\frac {47\,a^3}{12}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {29\,a^3}{4}+\frac {249\,a\,b^2}{32}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {29\,a^3}{4}+\frac {249\,a\,b^2}{32}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {435\,a\,b^2}{32}-\frac {33\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {435\,a\,b^2}{32}-\frac {33\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (12\,a^2\,b-4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {12\,a^2\,b}{7}-\frac {4\,b^3}{7}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (36\,a^2\,b-12\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (48\,a^2\,b+12\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (36\,a^2\,b+\frac {20\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (60\,a^2\,b-20\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {132\,a^2\,b}{7}+\frac {12\,b^3}{7}\right )-\frac {4\,b^3}{63}+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (atanh(tan(c/2 + (d*x)/2))*((15*a*b^2)/64 - (5*a^3)/8))/d - (tan(c/2 + (d*x)/2)*((15*a*b^2)/64 + (11*a^3)/8)
+ (6*a^2*b)/7 - tan(c/2 + (d*x)/2)^17*((15*a*b^2)/64 + (11*a^3)/8) + tan(c/2 + (d*x)/2)^3*((191*a*b^2)/32 - (
47*a^3)/12) - tan(c/2 + (d*x)/2)^15*((191*a*b^2)/32 - (47*a^3)/12) + tan(c/2 + (d*x)/2)^5*((249*a*b^2)/32 + (2
9*a^3)/4) - tan(c/2 + (d*x)/2)^13*((249*a*b^2)/32 + (29*a^3)/4) + tan(c/2 + (d*x)/2)^7*((435*a*b^2)/32 - (33*a
^3)/4) - tan(c/2 + (d*x)/2)^11*((435*a*b^2)/32 - (33*a^3)/4) - tan(c/2 + (d*x)/2)^14*(12*a^2*b - 4*b^3) - tan(
c/2 + (d*x)/2)^2*((12*a^2*b)/7 - (4*b^3)/7) - tan(c/2 + (d*x)/2)^6*(36*a^2*b - 12*b^3) + tan(c/2 + (d*x)/2)^8*
(48*a^2*b + 12*b^3) + tan(c/2 + (d*x)/2)^12*(36*a^2*b + (20*b^3)/3) - tan(c/2 + (d*x)/2)^10*(60*a^2*b - 20*b^3
) + tan(c/2 + (d*x)/2)^4*((132*a^2*b)/7 + (12*b^3)/7) - (4*b^3)/63 + 6*a^2*b*tan(c/2 + (d*x)/2)^16)/(d*(9*tan(
c/2 + (d*x)/2)^2 - 36*tan(c/2 + (d*x)/2)^4 + 84*tan(c/2 + (d*x)/2)^6 - 126*tan(c/2 + (d*x)/2)^8 + 126*tan(c/2
+ (d*x)/2)^10 - 84*tan(c/2 + (d*x)/2)^12 + 36*tan(c/2 + (d*x)/2)^14 - 9*tan(c/2 + (d*x)/2)^16 + tan(c/2 + (d*x
)/2)^18 - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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