3.51 \(\int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=67 \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b \sec (c+d x)}{d}-\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0936005, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b \sec (c+d x)}{d}-\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.0469392, size = 67, normalized size = 1. \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b \sec (c+d x)}{d}-\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.102, size = 98, normalized size = 1.5 \begin{align*}{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{ab}{d\cos \left ( dx+c \right ) }}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,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))^2,x)

[Out]

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

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Maxima [A]  time = 1.19866, size = 120, normalized size = 1.79 \begin{align*} -\frac{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 )} - 2 \, a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac{8 \, a b}{\cos \left (d x + c\right )}}{4 \, 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))^2,x, algorithm="maxima")

[Out]

-1/4*(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)) - 2*a^2*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 8*a*b/cos(d*x + c))/d

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Fricas [A]  time = 0.498405, size = 234, normalized size = 3.49 \begin{align*} \frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b \cos \left (d x + c\right ) + 2 \, b^{2} \sin \left (d x + c\right )}{4 \, 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))^2,x, algorithm="fricas")

[Out]

1/4*((2*a^2 - b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^2 - b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)
+ 8*a*b*cos(d*x + c) + 2*b^2*sin(d*x + c))/(d*cos(d*x + c)^2)

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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)**3*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.18976, size = 165, normalized size = 2.46 \begin{align*} \frac{{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a b\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{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))^2,x, algorithm="giac")

[Out]

1/2*((2*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(
b^2*tan(1/2*d*x + 1/2*c)^3 - 4*a*b*tan(1/2*d*x + 1/2*c)^2 + b^2*tan(1/2*d*x + 1/2*c) + 4*a*b)/(tan(1/2*d*x + 1
/2*c)^2 - 1)^2)/d