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

Optimal. Leaf size=72 \[ -\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac{2 a b^2 \tan (c+d x)}{d}+\frac{b (a+b \tan (c+d x))^2}{2 d} \]

[Out]

a*(a^2 - 3*b^2)*x - (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/d + (2*a*b^2*Tan[c + d*x])/d + (b*(a + b*Tan[c + d*x])
^2)/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0925312, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3086, 3482, 3525, 3475} \[ -\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac{2 a b^2 \tan (c+d x)}{d}+\frac{b (a+b \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*(a^2 - 3*b^2)*x - (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/d + (2*a*b^2*Tan[c + d*x])/d + (b*(a + b*Tan[c + d*x])
^2)/(2*d)

Rule 3086

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(a + b*Tan[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [C]  time = 0.267994, size = 79, normalized size = 1.1 \[ \frac{6 a b^2 \tan (c+d x)+(-b+i a)^3 \log (-\tan (c+d x)+i)-(b+i a)^3 \log (\tan (c+d x)+i)+b^3 \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a - b)^3*Log[I - Tan[c + d*x]] - (I*a + b)^3*Log[I + Tan[c + d*x]] + 6*a*b^2*Tan[c + d*x] + b^3*Tan[c + d*
x]^2)/(2*d)

________________________________________________________________________________________

Maple [A]  time = 0.122, size = 93, normalized size = 1.3 \begin{align*}{a}^{3}x+{\frac{{a}^{3}c}{d}}-3\,{\frac{{a}^{2}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-3\,a{b}^{2}x+3\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}-3\,{\frac{a{b}^{2}c}{d}}+{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.7547, size = 115, normalized size = 1.6 \begin{align*} \frac{2 \,{\left (d x + c\right )} a^{3} - 6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{3}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 3 \, a^{2} b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*a^3 - 6*(d*x + c - tan(d*x + c))*a*b^2 - b^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1
)) - 3*a^2*b*log(-sin(d*x + c)^2 + 1))/d

________________________________________________________________________________________

Fricas [A]  time = 0.505707, size = 215, normalized size = 2.99 \begin{align*} \frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + b^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.20417, size = 96, normalized size = 1.33 \begin{align*} \frac{b^{3} \tan \left (d x + c\right )^{2} + 6 \, a b^{2} \tan \left (d x + c\right ) + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )} +{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*tan(d*x + c)^2 + 6*a*b^2*tan(d*x + c) + 2*(a^3 - 3*a*b^2)*(d*x + c) + (3*a^2*b - b^3)*log(tan(d*x + c
)^2 + 1))/d