3.66 \(\int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=103 \[ \frac{3 a^2 b \sec (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 a b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a b^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{b^3 \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.122525, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3090, 3770, 2606, 8, 2611} \[ \frac{3 a^2 b \sec (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 a b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a b^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{b^3 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

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

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]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=\int \left (a^3 \sec (c+d x)+3 a^2 b \sec (c+d x) \tan (c+d x)+3 a b^2 \sec (c+d x) \tan ^2(c+d x)+b^3 \sec (c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sec (c+d x) \, dx+\left (3 a^2 b\right ) \int \sec (c+d x) \tan (c+d x) \, dx+\left (3 a b^2\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sec (c+d x) \tan ^3(c+d x) \, dx\\ &=\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a b^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \left (3 a b^2\right ) \int \sec (c+d x) \, dx+\frac{\left (3 a^2 b\right ) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+d x))}{d}+\frac{b^3 \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 a b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a^2 b \sec (c+d x)}{d}-\frac{b^3 \sec (c+d x)}{d}+\frac{b^3 \sec ^3(c+d x)}{3 d}+\frac{3 a b^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 1.5871, size = 293, normalized size = 2.84 \[ \frac{-6 a \left (2 a^2-3 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 b \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\left (18 a^2-5 b^2\right ) \cos (2 (c+d x))+18 a^2+2 b^2 \cos (c+d x)-b^2\right )+36 a^2 b+12 a^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{9 a b^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{9 a b^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-18 a b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-10 b^3}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(36*a^2*b - 10*b^3 - 6*a*(2*a^2 - 3*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^3*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] - 18*a*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (9*a*b^2)/(Cos[(c + d*x)/2] - Sin[
(c + d*x)/2])^2 + b^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 2*b*(18*a^2 - b^2 + 2*b^2*Cos[c + d*x] + (18*a
^2 - 5*b^2)*Cos[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]^2 - (9*a*b^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^2 + b^3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(12*d)

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Maple [A]  time = 0.124, size = 187, normalized size = 1.8 \begin{align*}{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}b}{d\cos \left ( dx+c \right ) }}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{3\,d}}-{\frac{2\,{b}^{3}\cos \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.24014, size = 159, normalized size = 1.54 \begin{align*} -\frac{9 \, a 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 )} - 6 \, a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac{36 \, a^{2} b}{\cos \left (d x + c\right )} + \frac{4 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/12*(9*a*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)) - 6*a^3*(
log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*a^2*b/cos(d*x + c) + 4*(3*cos(d*x + c)^2 - 1)*b^3/cos(d*x
+ c)^3)/d

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Fricas [A]  time = 0.510289, size = 304, normalized size = 2.95 \begin{align*} \frac{3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 18 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b^{3} + 12 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(3*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3*log(-sin(d
*x + c) + 1) + 18*a*b^2*cos(d*x + c)*sin(d*x + c) + 4*b^3 + 12*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)
^3)

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

[Out]

Timed out

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Giac [A]  time = 1.20761, size = 231, normalized size = 2.24 \begin{align*} \frac{3 \,{\left (2 \, 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 ) - 3 \,{\left (2 \, 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 (9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 18 \, a^{2} b + 4 \, b^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(3*(2*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b*tan(1/2*d*x + 1/2*c)^2
- 12*b^3*tan(1/2*d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d*x + 1/2*c) - 18*a^2*b + 4*b^3)/(tan(1/2*d*x + 1/2*c)^2 -
1)^3)/d