3.306 \(\int \cos ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=209 \[ -\frac{b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \tan (c+d x)}{2 d}+\frac{b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{1}{2} a^2 x \left (a^2 A+8 a b B+12 A b^2\right )+\frac{a (2 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]

[Out]

(a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*x)/2 + (b^2*(8*a*A*b + 12*a^2*B + b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(
5*A*b + 2*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (a*A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d
*x])/(2*d) - (b*(13*a^2*A*b - 2*A*b^3 + 4*a^3*B - 8*a*b^2*B)*Tan[c + d*x])/(2*d) - (b^2*(6*a*A*b + 2*a^2*B - b
^2*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.462845, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4025, 4094, 4048, 3770, 3767, 8} \[ -\frac{b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \tan (c+d x)}{2 d}+\frac{b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{1}{2} a^2 x \left (a^2 A+8 a b B+12 A b^2\right )+\frac{a (2 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*x)/2 + (b^2*(8*a*A*b + 12*a^2*B + b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(
5*A*b + 2*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (a*A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d
*x])/(2*d) - (b*(13*a^2*A*b - 2*A*b^3 + 4*a^3*B - 8*a*b^2*B)*Tan[c + d*x])/(2*d) - (b^2*(6*a*A*b + 2*a^2*B - b
^2*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4094

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + 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, B, 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 ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (-a (5 A b+2 a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \sec (c+d x)+2 b (a A-b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{1}{2} \int (a+b \sec (c+d x)) \left (-a \left (a^2 A+12 A b^2+8 a b B\right )+b \left (a^2 A-2 A b^2-6 a b B\right ) \sec (c+d x)+2 b \left (6 a A b+2 a^2 B-b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{4} \int \left (-2 a^2 \left (a^2 A+12 A b^2+8 a b B\right )-2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \sec (c+d x)+2 b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (b^2 \left (8 a A b+12 a^2 B+b^2 B\right )\right ) \int \sec (c+d x) \, dx-\frac{1}{2} \left (b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac{b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\left (b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=\frac{1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac{b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \tan (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.89767, size = 310, normalized size = 1.48 \[ \frac{2 a^2 (c+d x) \left (a^2 A+8 a b B+12 A b^2\right )-2 b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a^3 (a B+4 A b) \sin (c+d x)+a^4 A \sin (2 (c+d x))+\frac{4 b^3 (4 a B+A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b^3 (4 a B+A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{b^4 B}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^4 B}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*(c + d*x) - 2*b^2*(8*a*A*b + 12*a^2*B + b^2*B)*Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] + 2*b^2*(8*a*A*b + 12*a^2*B + b^2*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^4*B)/(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b^3*(A*b + 4*a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
- (b^4*B)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b^3*(A*b + 4*a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]) + 4*a^3*(4*A*b + a*B)*Sin[c + d*x] + a^4*A*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.064, size = 236, normalized size = 1.1 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}Ax}{2}}+{\frac{A{a}^{4}c}{2\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{3}b\sin \left ( dx+c \right ) }{d}}+4\,B{a}^{3}bx+4\,{\frac{B{a}^{3}bc}{d}}+6\,A{a}^{2}{b}^{2}x+6\,{\frac{A{a}^{2}{b}^{2}c}{d}}+6\,{\frac{B{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Ba{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{b}^{4}\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)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

1/2/d*A*a^4*sin(d*x+c)*cos(d*x+c)+1/2*a^4*A*x+1/2/d*A*a^4*c+1/d*B*a^4*sin(d*x+c)+4/d*A*a^3*b*sin(d*x+c)+4*B*a^
3*b*x+4/d*B*a^3*b*c+6*A*a^2*b^2*x+6/d*A*a^2*b^2*c+6/d*B*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4/d*A*a*b^3*ln(sec(d
*x+c)+tan(d*x+c))+4/d*B*a*b^3*tan(d*x+c)+1/d*A*b^4*tan(d*x+c)+1/2/d*B*b^4*sec(d*x+c)*tan(d*x+c)+1/2/d*B*b^4*ln
(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.981407, size = 282, normalized size = 1.35 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 16 \,{\left (d x + c\right )} B a^{3} b + 24 \,{\left (d x + c\right )} A a^{2} b^{2} - B b^{4}{\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 )} + 12 \, B a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{4} \sin \left (d x + c\right ) + 16 \, A a^{3} b \sin \left (d x + c\right ) + 16 \, B a b^{3} \tan \left (d x + c\right ) + 4 \, A b^{4} \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)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 16*(d*x + c)*B*a^3*b + 24*(d*x + c)*A*a^2*b^2 - B*b^4*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*B*a^2*b^2*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 8*A*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a^4*sin(d*x + c
) + 16*A*a^3*b*sin(d*x + c) + 16*B*a*b^3*tan(d*x + c) + 4*A*b^4*tan(d*x + c))/d

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Fricas [A]  time = 0.60597, size = 479, normalized size = 2.29 \begin{align*} \frac{2 \,{\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{2} +{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{4} \cos \left (d x + c\right )^{3} + B b^{4} + 2 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\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)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(2*(A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*d*x*cos(d*x + c)^2 + (12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*cos(d*x + c)
^2*log(sin(d*x + c) + 1) - (12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4
*cos(d*x + c)^3 + B*b^4 + 2*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^2 + 2*(4*B*a*b^3 + A*b^4)*cos(d*x + c))*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)**2*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.30352, size = 713, normalized size = 3.41 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/2*((A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*(d*x + c) + (12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*log(abs(tan(1/2*d*x + 1
/2*c) + 1)) - (12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(A*a^4*tan(1/2*d*x + 1
/2*c)^7 - 2*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^7
+ 2*A*b^4*tan(1/2*d*x + 1/2*c)^7 - B*b^4*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 2*B*a^4*ta
n(1/2*d*x + 1/2*c)^5 + 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 2*A*b^4*tan(1/2*d
*x + 1/2*c)^5 - 3*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^4*tan(1/2*d*x + 1/2*c)
^3 + 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*
B*b^4*tan(1/2*d*x + 1/2*c)^3 - A*a^4*tan(1/2*d*x + 1/2*c) - 2*B*a^4*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b*tan(1/2*d
*x + 1/2*c) - 8*B*a*b^3*tan(1/2*d*x + 1/2*c) - 2*A*b^4*tan(1/2*d*x + 1/2*c) - B*b^4*tan(1/2*d*x + 1/2*c))/(tan
(1/2*d*x + 1/2*c)^4 - 1)^2)/d