3.536 \(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.312713, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3048, 3031, 3023, 2735, 3770} \[ \frac{\left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A b \tan (c+d x)}{d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+2 a b C x-\frac{b^2 (A-2 C) \sin (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

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

Rubi steps

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

Mathematica [B]  time = 1.28264, size = 249, normalized size = 2.42 \[ \frac{-2 \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^2 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{8 a A b \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{8 a A b \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+8 a b C (c+d x)+4 b^2 C \sin (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*b*C*(c + d*x) - 2*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 + a^2*(
A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (8*a*A*
b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (a^2*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 +
(8*a*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*b^2*C*Sin[c + d*x])/(4*d)

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Maple [A]  time = 0.053, size = 133, normalized size = 1.3 \begin{align*}{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}C\sin \left ( dx+c \right ) }{d}}+2\,{\frac{aAb\tan \left ( dx+c \right ) }{d}}+2\,abCx+2\,{\frac{Cabc}{d}}+{\frac{A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

[Out]

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

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Maxima [A]  time = 1.0088, size = 189, normalized size = 1.83 \begin{align*} \frac{8 \,{\left (d x + c\right )} C a b - A a^{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 \, C b^{2} \sin \left (d x + c\right ) + 8 \, A a b \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60001, size = 347, normalized size = 3.37 \begin{align*} \frac{8 \, C a b d x \cos \left (d x + c\right )^{2} +{\left ({\left (A + 2 \, C\right )} a^{2} + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (A + 2 \, C\right )} a^{2} + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 4 \, A a b \cos \left (d x + c\right ) + A a^{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((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(8*C*a*b*d*x*cos(d*x + c)^2 + ((A + 2*C)*a^2 + 2*A*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - ((A + 2*C)*
a^2 + 2*A*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*b^2*cos(d*x + c)^2 + 4*A*a*b*cos(d*x + c) + A*a^
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((a+b*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.2545, size = 255, normalized size = 2.48 \begin{align*} \frac{4 \,{\left (d x + c\right )} C a b + \frac{4 \, C b^{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 (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (A a^{2} + 2 \, C a^{2} + 2 \, 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 (A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, A a b \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((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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