3.649 \(\int \cos (c+d x) (a+b \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=109 \[ \frac{\left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{2 a b (A-C) \tan (c+d x)}{d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^2}{d}+2 a A b x-\frac{b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.164549, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4095, 4048, 3770, 3767, 8} \[ \frac{\left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{2 a b (A-C) \tan (c+d x)}{d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^2}{d}+2 a A b x-\frac{b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4095

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e +
f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [B]  time = 0.87785, size = 352, normalized size = 3.23 \[ \frac{\sec ^2(c+d x) \left (\left (a^2 A+2 b^2 C\right ) \sin (c+d x)+\cos (2 (c+d x)) \left (-\left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+\left (2 a^2 C+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a A b (c+d x)\right )+a^2 A \sin (3 (c+d x))-2 a^2 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^2 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a A b c+4 a A b d x+4 a b C \sin (2 (c+d x))-2 A b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 A b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-b^2 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+b^2 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(4*a*A*b*c + 4*a*A*b*d*x - 2*A*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*a^2*C*Log[Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]] - b^2*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*A*b^2*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]] + 2*a^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^2*C*Log[Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]] + Cos[2*(c + d*x)]*(4*a*A*b*(c + d*x) - (2*A*b^2 + (2*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + (2*A*b^2 + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (a^2*A + 2*b^2*C)*Sin[c
+ d*x] + 4*a*b*C*Sin[2*(c + d*x)] + a^2*A*Sin[3*(c + d*x)]))/(4*d)

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Maple [A]  time = 0.063, size = 133, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,aAbx+2\,{\frac{Aabc}{d}}+2\,{\frac{abC\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}C\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(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*a^2*A*sin(d*x+c)+1/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+2*a*A*b*x+2/d*A*a*b*c+2/d*a*b*C*tan(d*x+c)+1/d*A*b^2*
ln(sec(d*x+c)+tan(d*x+c))+1/2/d*b^2*C*sec(d*x+c)*tan(d*x+c)+1/2/d*b^2*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.989743, size = 189, normalized size = 1.73 \begin{align*} \frac{8 \,{\left (d x + c\right )} A a b - C 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 \, C a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, C a b \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.537493, size = 347, normalized size = 3.18 \begin{align*} \frac{8 \, A a b d x \cos \left (d x + c\right )^{2} +{\left (2 \, C a^{2} +{\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, C a^{2} +{\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 4 \, C a b \cos \left (d x + c\right ) + C b^{2}\right )} \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(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.17603, size = 258, normalized size = 2.37 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a b + \frac{4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} +{\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (4 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\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(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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