3.103 \(\int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=224 \[ \frac {a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^4 b \sec (c+d x)}{d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {15 a b^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^5 \sec ^5(c+d x)}{5 d}-\frac {2 b^5 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d} \]

[Out]

a^5*arctanh(sin(d*x+c))/d-5*a^3*b^2*arctanh(sin(d*x+c))/d+15/8*a*b^4*arctanh(sin(d*x+c))/d+5*a^4*b*sec(d*x+c)/
d-10*a^2*b^3*sec(d*x+c)/d+b^5*sec(d*x+c)/d+10/3*a^2*b^3*sec(d*x+c)^3/d-2/3*b^5*sec(d*x+c)^3/d+1/5*b^5*sec(d*x+
c)^5/d+5*a^3*b^2*sec(d*x+c)*tan(d*x+c)/d-15/8*a*b^4*sec(d*x+c)*tan(d*x+c)/d+5/4*a*b^4*sec(d*x+c)*tan(d*x+c)^3/
d

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Rubi [A]  time = 0.23, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3090, 3770, 2606, 8, 2611, 194} \[ \frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \sec (c+d x)}{d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac {5 a^4 b \sec (c+d x)}{d}+\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {15 a b^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^5 \sec ^5(c+d x)}{5 d}-\frac {2 b^5 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && 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 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 ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \sec (c+d x)+5 a^4 b \sec (c+d x) \tan (c+d x)+10 a^3 b^2 \sec (c+d x) \tan ^2(c+d x)+10 a^2 b^3 \sec (c+d x) \tan ^3(c+d x)+5 a b^4 \sec (c+d x) \tan ^4(c+d x)+b^5 \sec (c+d x) \tan ^5(c+d x)\right ) \, dx\\ &=a^5 \int \sec (c+d x) \, dx+\left (5 a^4 b\right ) \int \sec (c+d x) \tan (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sec (c+d x) \tan ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sec (c+d x) \tan ^4(c+d x) \, dx+b^5 \int \sec (c+d x) \tan ^5(c+d x) \, dx\\ &=\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}+\frac {5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d}-\left (5 a^3 b^2\right ) \int \sec (c+d x) \, dx-\frac {1}{4} \left (15 a b^4\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\frac {\left (5 a^4 b\right ) \operatorname {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d}+\frac {\left (10 a^2 b^3\right ) \operatorname {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac {b^5 \operatorname {Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^4 b \sec (c+d x)}{d}-\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}+\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac {15 a b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {1}{8} \left (15 a b^4\right ) \int \sec (c+d x) \, dx+\frac {b^5 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a^5 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^4 b \sec (c+d x)}{d}-\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d}+\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac {2 b^5 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec ^5(c+d x)}{5 d}+\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac {15 a b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [B]  time = 6.29, size = 1219, normalized size = 5.44 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(b*(600*a^4 - 1000*a^2*b^2 + 89*b^4)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(120*d*(a*Cos[c + d*x] + b*Sin[c +
d*x])^5) + ((-8*a^5 + 40*a^3*b^2 - 15*a*b^4)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*T
an[c + d*x])^5)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((8*a^5 - 40*a^3*b^2 + 15*a*b^4)*Cos[c + d*x]^5*Lo
g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((2
5*a*b^4 + 2*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(80*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^5) + ((600*a^3*b^2 + 200*a^2*b^3 - 375*a*b^4 - 31*b^5)*Cos[c + d*x]^5*(a + b*Tan[c +
d*x])^5)/(240*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)/(20*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(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)/(20*d*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-25*a*b^4 + 2*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d
*x])^5)/(80*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-600*a^3*b^2 +
200*a^2*b^3 + 375*a*b^4 - 31*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(240*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*(-600*a^4*b*Sin[(c + d*x)/2] + 1000*a^2*b^3*S
in[(c + d*x)/2] - 89*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(120*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*(200*a^2*b^3*Sin[(c + d*x)/2] - 31*b^5*Sin[(c + d*x)/
2])*(a + b*Tan[c + d*x])^5)/(120*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*(-200*a^2*b^3*Sin[(c + d*x)/2] + 31*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(120*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*(600*a^4*b*Sin[(
c + d*x)/2] - 1000*a^2*b^3*Sin[(c + d*x)/2] + 89*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(120*d*(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5)

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fricas [A]  time = 0.64, size = 196, normalized size = 0.88 \[ \frac {15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, b^{5} + 240 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 160 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 150 \, {\left (2 \, a b^{4} \cos \left (d x + c\right ) + {\left (8 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

1/240*(15*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(8*a^5 - 40*a^3*b^2 + 15*a
*b^4)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 48*b^5 + 240*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 + 160*(
5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 150*(2*a*b^4*cos(d*x + c) + (8*a^3*b^2 - 5*a*b^4)*cos(d*x + c)^3)*sin(d*x +
c))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.64, size = 410, normalized size = 1.83 \[ \frac {15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (600 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 600 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 1200 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1050 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2400 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2400 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3600 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5600 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 640 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1200 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1050 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2400 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4000 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 320 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 600 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 600 \, a^{4} b + 800 \, a^{2} b^{3} - 64 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/120*(15*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*a^5 - 40*a^3*b^2 + 15*a*b
^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(600*a^3*b^2*tan(1/2*d*x + 1/2*c)^9 - 225*a*b^4*tan(1/2*d*x + 1/2*c
)^9 - 600*a^4*b*tan(1/2*d*x + 1/2*c)^8 - 1200*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 + 1050*a*b^4*tan(1/2*d*x + 1/2*c)
^7 + 2400*a^4*b*tan(1/2*d*x + 1/2*c)^6 - 2400*a^2*b^3*tan(1/2*d*x + 1/2*c)^6 - 3600*a^4*b*tan(1/2*d*x + 1/2*c)
^4 + 5600*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 640*b^5*tan(1/2*d*x + 1/2*c)^4 + 1200*a^3*b^2*tan(1/2*d*x + 1/2*c)^
3 - 1050*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 2400*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 4000*a^2*b^3*tan(1/2*d*x + 1/2*c)^
2 + 320*b^5*tan(1/2*d*x + 1/2*c)^2 - 600*a^3*b^2*tan(1/2*d*x + 1/2*c) + 225*a*b^4*tan(1/2*d*x + 1/2*c) - 600*a
^4*b + 800*a^2*b^3 - 64*b^5)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [B]  time = 0.23, size = 440, normalized size = 1.96 \[ \frac {a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {5 a^{4} b}{d \cos \left (d x +c \right )}+\frac {5 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {5 a^{3} b^{2} \sin \left (d x +c \right )}{d}-\frac {5 a^{3} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {10 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {10 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}-\frac {10 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{3 d}-\frac {20 a^{2} b^{3} \cos \left (d x +c \right )}{3 d}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {5 a \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 a \,b^{4} \sin \left (d x +c \right )}{8 d}+\frac {15 a \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {8 b^{5} \cos \left (d x +c \right )}{15 d}+\frac {b^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

1/d*a^5*ln(sec(d*x+c)+tan(d*x+c))+5/d*a^4*b/cos(d*x+c)+5/d*a^3*b^2*sin(d*x+c)^3/cos(d*x+c)^2+5*a^3*b^2*sin(d*x
+c)/d-5/d*a^3*b^2*ln(sec(d*x+c)+tan(d*x+c))+10/3/d*a^2*b^3*sin(d*x+c)^4/cos(d*x+c)^3-10/3/d*a^2*b^3*sin(d*x+c)
^4/cos(d*x+c)-10/3/d*cos(d*x+c)*sin(d*x+c)^2*a^2*b^3-20/3*a^2*b^3*cos(d*x+c)/d+5/4/d*a*b^4*sin(d*x+c)^5/cos(d*
x+c)^4-5/8/d*a*b^4*sin(d*x+c)^5/cos(d*x+c)^2-5/8*a*b^4*sin(d*x+c)^3/d-15/8*a*b^4*sin(d*x+c)/d+15/8/d*a*b^4*ln(
sec(d*x+c)+tan(d*x+c))+1/5/d*b^5*sin(d*x+c)^6/cos(d*x+c)^5-1/15/d*b^5*sin(d*x+c)^6/cos(d*x+c)^3+1/5/d*b^5*sin(
d*x+c)^6/cos(d*x+c)+8/15*b^5*cos(d*x+c)/d+1/5/d*b^5*cos(d*x+c)*sin(d*x+c)^4+4/15/d*cos(d*x+c)*sin(d*x+c)^2*b^5

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maxima [A]  time = 0.34, size = 230, normalized size = 1.03 \[ \frac {75 \, a b^{4} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 600 \, a^{3} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, a^{5} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {1200 \, a^{4} b}{\cos \left (d x + c\right )} - \frac {800 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac {16 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} b^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

1/240*(75*a*b^4*(2*(5*sin(d*x + c)^3 - 3*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)) - 600*a^3*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 120*a^5*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 1200*a^4*b/cos(d*x + c
) - 800*(3*cos(d*x + c)^2 - 1)*a^2*b^3/cos(d*x + c)^3 + 16*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 3)*b^5/cos
(d*x + c)^5)/d

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mupad [B]  time = 4.26, size = 345, normalized size = 1.54 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^5-10\,a^3\,b^2+\frac {15\,a\,b^4}{4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {15\,a\,b^4}{4}-10\,a^3\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {35\,a\,b^4}{2}-20\,a^3\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {35\,a\,b^4}{2}-20\,a^3\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (40\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (40\,a^4\,b-\frac {200\,a^2\,b^3}{3}+\frac {16\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (60\,a^4\,b-\frac {280\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+10\,a^4\,b+\frac {16\,b^5}{15}-\frac {40\,a^2\,b^3}{3}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {15\,a\,b^4}{4}-10\,a^3\,b^2\right )+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

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)^6,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((15*a*b^4)/4 + 2*a^5 - 10*a^3*b^2))/d - (tan(c/2 + (d*x)/2)^9*((15*a*b^4)/4 - 10*a
^3*b^2) + tan(c/2 + (d*x)/2)^3*((35*a*b^4)/2 - 20*a^3*b^2) - tan(c/2 + (d*x)/2)^7*((35*a*b^4)/2 - 20*a^3*b^2)
- tan(c/2 + (d*x)/2)^6*(40*a^4*b - 40*a^2*b^3) - tan(c/2 + (d*x)/2)^2*(40*a^4*b + (16*b^5)/3 - (200*a^2*b^3)/3
) + tan(c/2 + (d*x)/2)^4*(60*a^4*b + (32*b^5)/3 - (280*a^2*b^3)/3) + 10*a^4*b + (16*b^5)/15 - (40*a^2*b^3)/3 -
tan(c/2 + (d*x)/2)*((15*a*b^4)/4 - 10*a^3*b^2) + 10*a^4*b*tan(c/2 + (d*x)/2)^8)/(d*(5*tan(c/2 + (d*x)/2)^2 -
10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 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)**6*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

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