3.68 \(\int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=158 \[ \frac {a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a^2 b \sec ^3(c+d x)}{d}-\frac {3 a b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {3 a b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^3 \sec ^5(c+d x)}{5 d}-\frac {b^3 \sec ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 12, 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} \[ \frac {a^2 b \sec ^3(c+d x)}{d}+\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {3 a b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {3 a b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^3 \sec ^5(c+d x)}{5 d}-\frac {b^3 \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

________________________________________________________________________________________

Mathematica [B]  time = 1.32, size = 464, normalized size = 2.94 \[ \frac {\sec ^5(c+d x) \left (240 a^3 \sin (2 (c+d x))+120 a^3 \sin (4 (c+d x))-300 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-60 a^3 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+300 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+60 a^3 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+320 \left (3 a^2 b-b^3\right ) \cos (2 (c+d x))-150 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+960 a^2 b+540 a b^2 \sin (2 (c+d x))-90 a b^2 \sin (4 (c+d x))+225 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+45 a b^2 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-225 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-45 a b^2 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+64 b^3\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^5*(960*a^2*b + 64*b^3 + 320*(3*a^2*b - b^3)*Cos[2*(c + d*x)] - 300*a^3*Cos[3*(c + d*x)]*Log[Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]] + 225*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 60*a^
3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 45*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] - 150*a*(4*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]]) + 300*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 225*a*b^
2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 60*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] - 45*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 240*a^3*Sin[2*(c + d*x
)] + 540*a*b^2*Sin[2*(c + d*x)] + 120*a^3*Sin[4*(c + d*x)] - 90*a*b^2*Sin[4*(c + d*x)]))/(1920*d)

________________________________________________________________________________________

fricas [A]  time = 0.74, size = 147, normalized size = 0.93 \[ \frac {15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, b^{3} + 80 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) + {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

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))^3,x, algorithm="fricas")

[Out]

1/240*(15*(4*a^3 - 3*a*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*a^3 - 3*a*b^2)*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 48*b^3 + 80*(3*a^2*b - b^3)*cos(d*x + c)^2 + 30*(6*a*b^2*cos(d*x + c) + (4*a^3 - 3*a*b^2)*co
s(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

giac [B]  time = 0.52, size = 333, normalized size = 2.11 \[ \frac {15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 270 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 480 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 80 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 270 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 45 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, a^{2} b + 16 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/120*(15*(4*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) + 2*(60*a^3*tan(1/2*d*x + 1/2*c)^9 + 45*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 360*a^2*b*tan(1/2*d*x + 1/2*
c)^8 - 120*a^3*tan(1/2*d*x + 1/2*c)^7 + 270*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*a^2*b*tan(1/2*d*x + 1/2*c)^6 -
240*b^3*tan(1/2*d*x + 1/2*c)^6 - 480*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 80*b^3*tan(1/2*d*x + 1/2*c)^4 + 120*a^3*ta
n(1/2*d*x + 1/2*c)^3 - 270*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 240*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 80*b^3*tan(1/2*d*
x + 1/2*c)^2 - 60*a^3*tan(1/2*d*x + 1/2*c) - 45*a*b^2*tan(1/2*d*x + 1/2*c) - 120*a^2*b + 16*b^3)/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^5)/d

________________________________________________________________________________________

maple [A]  time = 2.36, size = 256, normalized size = 1.62 \[ \frac {a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{2} b}{d \cos \left (d x +c \right )^{3}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{8 d}-\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}-\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )}-\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}-\frac {2 b^{3} \cos \left (d x +c \right )}{15 d} \]

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))^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 157, normalized size = 0.99 \[ \frac {45 \, a b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {240 \, a^{2} b}{\cos \left (d x + c\right )^{3}} - \frac {16 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

1/240*(45*a*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)) - 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(
d*x + c) - 1)) + 240*a^2*b/cos(d*x + c)^3 - 16*(5*cos(d*x + c)^2 - 3)*b^3/cos(d*x + c)^5)/d

________________________________________________________________________________________

mupad [B]  time = 4.28, size = 293, normalized size = 1.85 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (a^3+\frac {3\,a\,b^2}{4}\right )-2\,a^2\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {9\,a\,b^2}{2}-2\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a\,b^2}{2}-2\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^2\,b-\frac {4\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^2\,b+\frac {4\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (12\,a^2\,b-4\,b^3\right )+\frac {4\,b^3}{15}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3+\frac {3\,a\,b^2}{4}\right )-6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a\,b^2}{4}-a^3\right )}{d} \]

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)^6,x)

[Out]

(tan(c/2 + (d*x)/2)^9*((3*a*b^2)/4 + a^3) - 2*a^2*b - tan(c/2 + (d*x)/2)^3*((9*a*b^2)/2 - 2*a^3) + tan(c/2 + (
d*x)/2)^7*((9*a*b^2)/2 - 2*a^3) + tan(c/2 + (d*x)/2)^2*(4*a^2*b - (4*b^3)/3) - tan(c/2 + (d*x)/2)^4*(8*a^2*b +
(4*b^3)/3) + tan(c/2 + (d*x)/2)^6*(12*a^2*b - 4*b^3) + (4*b^3)/15 - tan(c/2 + (d*x)/2)*((3*a*b^2)/4 + a^3) -
6*a^2*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)) - (atanh(tan(c/2 + (d*x)/2))*((3*a*b^2)/4 - a^3))/d

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________