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

Optimal. Leaf size=112 \[ -\frac{a^2 (2 A-3 C) \tan (c+d x)}{2 d}+\frac{a^2 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+2 a^2 A x+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.189976, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4087, 3917, 3914, 3767, 8, 3770} \[ -\frac{a^2 (2 A-3 C) \tan (c+d x)}{2 d}+\frac{a^2 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+2 a^2 A x+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4087

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

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]

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

Mathematica [B]  time = 2.61197, size = 330, normalized size = 2.95 \[ \frac{a^2 \cos ^4(c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{2 (2 A+3 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (2 A+3 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{4 A \sin (c) \cos (d x)}{d}+\frac{4 A \cos (c) \sin (d x)}{d}+8 A x+\frac{8 C \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{8 C \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{C}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{C}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{8 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[c + d*x]^4*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(8*A*x - (2*(2*A + 3*C)*Log
[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(2*A + 3*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*A*C
os[d*x]*Sin[c])/d + (4*A*Cos[c]*Sin[d*x])/d + C/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (8*C*Sin[(d*x)/2
])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - C/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
)^2) + (8*C*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(8*(A + 2*C + A*Co
s[2*(c + d*x)]))

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Maple [A]  time = 0.086, size = 114, normalized size = 1. \begin{align*}{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{a}^{2}Ax+2\,{\frac{A{a}^{2}c}{d}}+2\,{\frac{{a}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2*A*sin(d*x+c)+3/2/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+2*a^2*A*x+2/d*A*a^2*c+2/d*a^2*C*tan(d*x+c)+1/d*a^2*
A*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^2*C*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 0.938584, size = 192, normalized size = 1.71 \begin{align*} \frac{8 \,{\left (d x + c\right )} A a^{2} - C 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 \, A 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 \, 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 )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, C a^{2} \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+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.525572, size = 320, normalized size = 2.86 \begin{align*} \frac{8 \, A a^{2} d x \cos \left (d x + c\right )^{2} +{\left (2 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, A + 3 \, C\right )} a^{2} \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^{2} \cos \left (d x + c\right ) + C 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(cos(d*x+c)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.23061, size = 205, normalized size = 1.83 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a^{2} + \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 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, C a^{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+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/2*(4*(d*x + c)*A*a^2 + 4*A*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + (2*A*a^2 + 3*C*a^2)*log(a
bs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*C*a^2*tan(1/2*d*
x + 1/2*c)^3 - 5*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d