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

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

[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)

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Rubi [A]  time = 0.180935, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]]

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 = 1.31819, size = 464, normalized size = 2.94 \[ \frac{\sec ^5(c+d x) \left (320 \left (3 a^2 b-b^3\right ) \cos (2 (c+d x))-150 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+960 a^2 b+240 a^3 \sin (2 (c+d x))+120 a^3 \sin (4 (c+d x))-300 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-60 a^3 \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+300 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+60 a^3 \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+540 a b^2 \sin (2 (c+d x))-90 a b^2 \sin (4 (c+d x))+225 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+45 a b^2 \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-225 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-45 a b^2 \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+64 b^3\right )}{1920 d} \]

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)

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Maple [A]  time = 0.131, size = 256, normalized size = 1.6 \begin{align*}{\frac{{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{2}b}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{3\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,d\cos \left ( dx+c \right ) }}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{15\,d}}-{\frac{2\,{b}^{3}\cos \left ( dx+c \right ) }{15\,d}} \end{align*}

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*a*b^2*sin(d*x
+c)^3/cos(d*x+c)^4+3/8/d*a*b^2*sin(d*x+c)^3/cos(d*x+c)^2+3/8*a*b^2*sin(d*x+c)/d-3/8/d*a*b^2*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*cos(d*x+c)*sin(d*x+c)^2*b^3-2/15*b^3*cos(d*x+c)/d

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Maxima [A]  time = 1.2544, size = 212, normalized size = 1.34 \begin{align*} \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} \end{align*}

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

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Fricas [A]  time = 0.520492, size = 362, normalized size = 2.29 \begin{align*} \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}} \end{align*}

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)

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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)**6*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.23838, size = 450, normalized size = 2.85 \begin{align*} \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} \end{align*}

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