3.105 \(\int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=318 \[ \frac {a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^5 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {5 a^4 b \sec ^3(c+d x)}{3 d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {5 a^3 b^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}-\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{4 d}+\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}+\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {5 a b^4 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {5 a b^4 \tan (c+d x) \sec (c+d x)}{16 d}+\frac {b^5 \sec ^7(c+d x)}{7 d}-\frac {2 b^5 \sec ^5(c+d x)}{5 d}+\frac {b^5 \sec ^3(c+d x)}{3 d} \]

[Out]

1/2*a^5*arctanh(sin(d*x+c))/d-5/4*a^3*b^2*arctanh(sin(d*x+c))/d+5/16*a*b^4*arctanh(sin(d*x+c))/d+5/3*a^4*b*sec
(d*x+c)^3/d-10/3*a^2*b^3*sec(d*x+c)^3/d+1/3*b^5*sec(d*x+c)^3/d+2*a^2*b^3*sec(d*x+c)^5/d-2/5*b^5*sec(d*x+c)^5/d
+1/7*b^5*sec(d*x+c)^7/d+1/2*a^5*sec(d*x+c)*tan(d*x+c)/d-5/4*a^3*b^2*sec(d*x+c)*tan(d*x+c)/d+5/16*a*b^4*sec(d*x
+c)*tan(d*x+c)/d+5/2*a^3*b^2*sec(d*x+c)^3*tan(d*x+c)/d-5/8*a*b^4*sec(d*x+c)^3*tan(d*x+c)/d+5/6*a*b^4*sec(d*x+c
)^3*tan(d*x+c)^3/d

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Rubi [A]  time = 0.34, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 19, 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} \[ \frac {2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {5 a^3 b^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}-\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{4 d}+\frac {5 a^4 b \sec ^3(c+d x)}{3 d}+\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^5 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {5 a b^4 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {5 a b^4 \tan (c+d x) \sec (c+d x)}{16 d}+\frac {b^5 \sec ^7(c+d x)}{7 d}-\frac {2 b^5 \sec ^5(c+d x)}{5 d}+\frac {b^5 \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(a^5*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^3*b^2*ArcTanh[Sin[c + d*x]])/(4*d) + (5*a*b^4*ArcTanh[Sin[c + d*x]])/
(16*d) + (5*a^4*b*Sec[c + d*x]^3)/(3*d) - (10*a^2*b^3*Sec[c + d*x]^3)/(3*d) + (b^5*Sec[c + d*x]^3)/(3*d) + (2*
a^2*b^3*Sec[c + d*x]^5)/d - (2*b^5*Sec[c + d*x]^5)/(5*d) + (b^5*Sec[c + d*x]^7)/(7*d) + (a^5*Sec[c + d*x]*Tan[
c + d*x])/(2*d) - (5*a^3*b^2*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (5*a*b^4*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (
5*a^3*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) - (5*a*b^4*Sec[c + d*x]^3*Tan[c + d*x])/(8*d) + (5*a*b^4*Sec[c +
d*x]^3*Tan[c + d*x]^3)/(6*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 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 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 ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \sec ^3(c+d x)+5 a^4 b \sec ^3(c+d x) \tan (c+d x)+10 a^3 b^2 \sec ^3(c+d x) \tan ^2(c+d x)+10 a^2 b^3 \sec ^3(c+d x) \tan ^3(c+d x)+5 a b^4 \sec ^3(c+d x) \tan ^4(c+d x)+b^5 \sec ^3(c+d x) \tan ^5(c+d x)\right ) \, dx\\ &=a^5 \int \sec ^3(c+d x) \, dx+\left (5 a^4 b\right ) \int \sec ^3(c+d x) \tan (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sec ^3(c+d x) \tan ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx+b^5 \int \sec ^3(c+d x) \tan ^5(c+d x) \, dx\\ &=\frac {a^5 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac {1}{2} a^5 \int \sec (c+d x) \, dx-\frac {1}{2} \left (5 a^3 b^2\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{2} \left (5 a b^4\right ) \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx+\frac {\left (5 a^4 b\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (10 a^2 b^3\right ) \operatorname {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac {b^5 \operatorname {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^4 b \sec ^3(c+d x)}{3 d}+\frac {a^5 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}-\frac {5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}-\frac {1}{4} \left (5 a^3 b^2\right ) \int \sec (c+d x) \, dx+\frac {1}{8} \left (5 a b^4\right ) \int \sec ^3(c+d x) \, dx+\frac {\left (10 a^2 b^3\right ) \operatorname {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac {b^5 \operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {5 a^4 b \sec ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec ^3(c+d x)}{3 d}+\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac {2 b^5 \sec ^5(c+d x)}{5 d}+\frac {b^5 \sec ^7(c+d x)}{7 d}+\frac {a^5 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {5 a b^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}-\frac {5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac {1}{16} \left (5 a b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {5 a^4 b \sec ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec ^3(c+d x)}{3 d}+\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac {2 b^5 \sec ^5(c+d x)}{5 d}+\frac {b^5 \sec ^7(c+d x)}{7 d}+\frac {a^5 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {5 a b^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}-\frac {5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 6.34, size = 1677, normalized size = 5.27 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(b*(1400*a^4 - 1540*a^2*b^2 + 103*b^4)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(1680*d*(a*Cos[c + d*x] + b*Sin[
c + d*x])^5) + ((-8*a^5 + 20*a^3*b^2 - 5*a*b^4)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b
*Tan[c + d*x])^5)/(16*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((8*a^5 - 20*a^3*b^2 + 5*a*b^4)*Cos[c + d*x]^5*
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5)/(16*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) +
((35*a*b^4 + 3*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(336*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6*(a*C
os[c + d*x] + b*Sin[c + d*x])^5) + ((350*a^3*b^2 + 140*a^2*b^3 - 175*a*b^4 - 18*b^5)*Cos[c + d*x]^5*(a + b*Tan
[c + d*x])^5)/(560*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((840*a^5
+ 1400*a^4*b - 2100*a^3*b^2 - 1540*a^2*b^3 + 525*a*b^4 + 103*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(3360
*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (b^5*Cos[c + d*x]^5*Sin[(c +
d*x)/2]*(a + b*Tan[c + d*x])^5)/(56*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^7*(a*Cos[c + d*x] + b*Sin[c + d*x
])^5) - (b^5*Cos[c + d*x]^5*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^5)/(56*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-35*a*b^4 + 3*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(336*d
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-350*a^3*b^2 + 140*a^2*b^3 +
175*a*b^4 - 18*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(560*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(a*
Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-840*a^5 + 1400*a^4*b + 2100*a^3*b^2 - 1540*a^2*b^3 - 525*a*b^4 + 103*b^
5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(3360*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*
Sin[c + d*x])^5) + (Cos[c + d*x]^5*(-1400*a^4*b*Sin[(c + d*x)/2] + 1540*a^2*b^3*Sin[(c + d*x)/2] - 103*b^5*Sin
[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(1680*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin
[c + d*x])^5) + (Cos[c + d*x]^5*(-1400*a^4*b*Sin[(c + d*x)/2] + 1540*a^2*b^3*Sin[(c + d*x)/2] - 103*b^5*Sin[(c
+ d*x)/2])*(a + b*Tan[c + d*x])^5)/(1680*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c +
d*x])^5) + (Cos[c + d*x]^5*(70*a^2*b^3*Sin[(c + d*x)/2] - 9*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(140
*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (Cos[c + d*x]^5*(-70*a^2*b^3
*Sin[(c + d*x)/2] + 9*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(140*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (Cos[c + d*x]^5*(1400*a^4*b*Sin[(c + d*x)/2] - 1540*a^2*b^3*Sin[(c
+ d*x)/2] + 103*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(1680*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (Cos[c + d*x]^5*(1400*a^4*b*Sin[(c + d*x)/2] - 1540*a^2*b^3*Sin[(c + d
*x)/2] + 103*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(1680*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Co
s[c + d*x] + b*Sin[c + d*x])^5)

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fricas [A]  time = 0.59, size = 227, normalized size = 0.71 \[ \frac {105 \, {\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 480 \, b^{5} + 1120 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 1344 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 70 \, {\left (40 \, a b^{4} \cos \left (d x + c\right ) + 3 \, {\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 10 \, {\left (12 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3360 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

1/3360*(105*(8*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(8*a^5 - 20*a^3*b^2 + 5*
a*b^4)*cos(d*x + c)^7*log(-sin(d*x + c) + 1) + 480*b^5 + 1120*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 + 13
44*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 70*(40*a*b^4*cos(d*x + c) + 3*(8*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos(d*x + c
)^5 + 10*(12*a^3*b^2 - 7*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^7)

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giac [B]  time = 6.46, size = 680, normalized size = 2.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/1680*(105*(8*a^5 - 20*a^3*b^2 + 5*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(8*a^5 - 20*a^3*b^2 + 5*a*
b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(840*a^5*tan(1/2*d*x + 1/2*c)^13 + 2100*a^3*b^2*tan(1/2*d*x + 1/2*
c)^13 - 525*a*b^4*tan(1/2*d*x + 1/2*c)^13 - 8400*a^4*b*tan(1/2*d*x + 1/2*c)^12 - 3360*a^5*tan(1/2*d*x + 1/2*c)
^11 + 8400*a^3*b^2*tan(1/2*d*x + 1/2*c)^11 + 3500*a*b^4*tan(1/2*d*x + 1/2*c)^11 + 33600*a^4*b*tan(1/2*d*x + 1/
2*c)^10 - 33600*a^2*b^3*tan(1/2*d*x + 1/2*c)^10 + 4200*a^5*tan(1/2*d*x + 1/2*c)^9 - 23100*a^3*b^2*tan(1/2*d*x
+ 1/2*c)^9 + 16975*a*b^4*tan(1/2*d*x + 1/2*c)^9 - 53200*a^4*b*tan(1/2*d*x + 1/2*c)^8 + 56000*a^2*b^3*tan(1/2*d
*x + 1/2*c)^8 - 8960*b^5*tan(1/2*d*x + 1/2*c)^8 + 44800*a^4*b*tan(1/2*d*x + 1/2*c)^6 - 22400*a^2*b^3*tan(1/2*d
*x + 1/2*c)^6 - 4480*b^5*tan(1/2*d*x + 1/2*c)^6 - 4200*a^5*tan(1/2*d*x + 1/2*c)^5 + 23100*a^3*b^2*tan(1/2*d*x
+ 1/2*c)^5 - 16975*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 25200*a^4*b*tan(1/2*d*x + 1/2*c)^4 + 13440*a^2*b^3*tan(1/2*d
*x + 1/2*c)^4 - 2688*b^5*tan(1/2*d*x + 1/2*c)^4 + 3360*a^5*tan(1/2*d*x + 1/2*c)^3 - 8400*a^3*b^2*tan(1/2*d*x +
1/2*c)^3 - 3500*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 11200*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 15680*a^2*b^3*tan(1/2*d*x
+ 1/2*c)^2 + 896*b^5*tan(1/2*d*x + 1/2*c)^2 - 840*a^5*tan(1/2*d*x + 1/2*c) - 2100*a^3*b^2*tan(1/2*d*x + 1/2*c
) + 525*a*b^4*tan(1/2*d*x + 1/2*c) - 2800*a^4*b + 2240*a^2*b^3 - 128*b^5)/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

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maple [A]  time = 0.28, size = 564, normalized size = 1.77 \[ \frac {8 b^{5} \cos \left (d x +c \right )}{105 d}-\frac {5 a \,b^{4} \sin \left (d x +c \right )}{16 d}+\frac {5 a^{3} b^{2} \sin \left (d x +c \right )}{4 d}+\frac {a^{5} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}-\frac {4 a^{2} b^{3} \cos \left (d x +c \right )}{3 d}-\frac {2 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{3 d}-\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{105 d \cos \left (d x +c \right )^{3}}-\frac {5 a^{3} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {5 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}+\frac {2 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{4}}+\frac {a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {5 a \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}-\frac {5 a \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{48 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )}+\frac {b^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{35 d}+\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{105 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{5}}+\frac {5 a^{4} b}{3 d \cos \left (d x +c \right )^{3}}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{7}}-\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{2}}+\frac {5 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}+\frac {2 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{5}}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}-\frac {2 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

8/105*b^5*cos(d*x+c)/d-5/16*a*b^4*sin(d*x+c)/d-5/48*a*b^4*sin(d*x+c)^3/d+5/4*a^3*b^2*sin(d*x+c)/d+1/2*a^5*sec(
d*x+c)*tan(d*x+c)/d-4/3*a^2*b^3*cos(d*x+c)/d+1/35/d*b^5*sin(d*x+c)^6/cos(d*x+c)^5-5/4/d*a^3*b^2*ln(sec(d*x+c)+
tan(d*x+c))-1/105/d*b^5*sin(d*x+c)^6/cos(d*x+c)^3+5/4/d*a^3*b^2*sin(d*x+c)^3/cos(d*x+c)^2+2/3/d*a^2*b^3*sin(d*
x+c)^4/cos(d*x+c)^3+5/24/d*a*b^4*sin(d*x+c)^5/cos(d*x+c)^4-2/3/d*cos(d*x+c)*sin(d*x+c)^2*a^2*b^3+5/2/d*a^3*b^2
*sin(d*x+c)^3/cos(d*x+c)^4+2/d*a^2*b^3*sin(d*x+c)^4/cos(d*x+c)^5+5/6/d*a*b^4*sin(d*x+c)^5/cos(d*x+c)^6+1/2/d*a
^5*ln(sec(d*x+c)+tan(d*x+c))+5/16/d*a*b^4*ln(sec(d*x+c)+tan(d*x+c))+1/35/d*b^5*sin(d*x+c)^6/cos(d*x+c)+1/35/d*
b^5*cos(d*x+c)*sin(d*x+c)^4+4/105/d*cos(d*x+c)*sin(d*x+c)^2*b^5-2/3/d*a^2*b^3*sin(d*x+c)^4/cos(d*x+c)-5/48/d*a
*b^4*sin(d*x+c)^5/cos(d*x+c)^2+5/3/d*a^4*b/cos(d*x+c)^3+1/7/d*b^5*sin(d*x+c)^6/cos(d*x+c)^7

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maxima [A]  time = 0.34, size = 289, normalized size = 0.91 \[ -\frac {175 \, a b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} + 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2100 \, a^{3} b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 840 \, a^{5} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {5600 \, a^{4} b}{\cos \left (d x + c\right )^{3}} + \frac {2240 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a^{2} b^{3}}{\cos \left (d x + c\right )^{5}} - \frac {32 \, {\left (35 \, \cos \left (d x + c\right )^{4} - 42 \, \cos \left (d x + c\right )^{2} + 15\right )} b^{5}}{\cos \left (d x + c\right )^{7}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/3360*(175*a*b^4*(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)) - 2100*a^3*b^2*(2*(sin(d*x + c)
^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) +
840*a^5*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 5600*a^4*b/cos
(d*x + c)^3 + 2240*(5*cos(d*x + c)^2 - 3)*a^2*b^3/cos(d*x + c)^5 - 32*(35*cos(d*x + c)^4 - 42*cos(d*x + c)^2 +
15)*b^5/cos(d*x + c)^7)/d

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mupad [B]  time = 4.21, size = 514, normalized size = 1.62 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^5-\frac {5\,a^3\,b^2}{2}+\frac {5\,a\,b^4}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-4\,a^5+10\,a^3\,b^2+\frac {25\,a\,b^4}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (40\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (a^5+\frac {5\,a^3\,b^2}{2}-\frac {5\,a\,b^4}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (-4\,a^5+10\,a^3\,b^2+\frac {25\,a\,b^4}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (5\,a^5-\frac {55\,a^3\,b^2}{2}+\frac {485\,a\,b^4}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (5\,a^5-\frac {55\,a^3\,b^2}{2}+\frac {485\,a\,b^4}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (30\,a^4\,b-16\,a^2\,b^3+\frac {16\,b^5}{5}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {40\,a^4\,b}{3}-\frac {56\,a^2\,b^3}{3}+\frac {16\,b^5}{15}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-\frac {160\,a^4\,b}{3}+\frac {80\,a^2\,b^3}{3}+\frac {16\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {190\,a^4\,b}{3}-\frac {200\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+\frac {10\,a^4\,b}{3}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^5+\frac {5\,a^3\,b^2}{2}-\frac {5\,a\,b^4}{8}\right )+\frac {16\,b^5}{105}-\frac {8\,a^2\,b^3}{3}+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\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))^5/cos(c + d*x)^8,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((5*a*b^4)/8 + a^5 - (5*a^3*b^2)/2))/d - (tan(c/2 + (d*x)/2)^3*((25*a*b^4)/6 - 4*a^
5 + 10*a^3*b^2) - tan(c/2 + (d*x)/2)^10*(40*a^4*b - 40*a^2*b^3) - tan(c/2 + (d*x)/2)^13*(a^5 - (5*a*b^4)/8 + (
5*a^3*b^2)/2) - tan(c/2 + (d*x)/2)^11*((25*a*b^4)/6 - 4*a^5 + 10*a^3*b^2) + tan(c/2 + (d*x)/2)^5*((485*a*b^4)/
24 + 5*a^5 - (55*a^3*b^2)/2) - tan(c/2 + (d*x)/2)^9*((485*a*b^4)/24 + 5*a^5 - (55*a^3*b^2)/2) + tan(c/2 + (d*x
)/2)^4*(30*a^4*b + (16*b^5)/5 - 16*a^2*b^3) - tan(c/2 + (d*x)/2)^2*((40*a^4*b)/3 + (16*b^5)/15 - (56*a^2*b^3)/
3) + tan(c/2 + (d*x)/2)^6*((16*b^5)/3 - (160*a^4*b)/3 + (80*a^2*b^3)/3) + tan(c/2 + (d*x)/2)^8*((190*a^4*b)/3
+ (32*b^5)/3 - (200*a^2*b^3)/3) + (10*a^4*b)/3 + tan(c/2 + (d*x)/2)*(a^5 - (5*a*b^4)/8 + (5*a^3*b^2)/2) + (16*
b^5)/105 - (8*a^2*b^3)/3 + 10*a^4*b*tan(c/2 + (d*x)/2)^12)/(d*(7*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^
4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 - 7*tan(c/2 + (d*x)/2)^12 + t
an(c/2 + (d*x)/2)^14 - 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)**8*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

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