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

Optimal. Leaf size=160 \[ \frac{5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac{a^4 (8 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac{1}{2} a^4 x (13 A+8 B)+\frac{a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.389536, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4017, 4018, 3996, 3770} \[ \frac{5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac{a^4 (8 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac{1}{2} a^4 x (13 A+8 B)+\frac{a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4017

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

Rule 4018

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

Rule 3996

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

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 ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^3 (a (5 A+2 B)-2 a (A-B) \sec (c+d x)) \, dx\\ &=\frac{a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{1}{4} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (6 A+B)+2 a^2 (A+6 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{1}{4} \int \cos (c+d x) (a+a \sec (c+d x)) \left (10 a^3 (A-B)+2 a^3 (8 A+13 B) \sec (c+d x)\right ) \, dx\\ &=\frac{5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac{1}{4} \int \left (-2 a^4 (13 A+8 B)-2 a^4 (8 A+13 B) \sec (c+d x)\right ) \, dx\\ &=\frac{1}{2} a^4 (13 A+8 B) x+\frac{5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{1}{2} \left (a^4 (8 A+13 B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (13 A+8 B) x+\frac{a^4 (8 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 4.83511, size = 373, normalized size = 2.33 \[ \frac{a^4 \cos ^5(c+d x) \sec ^8\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 (A+B \sec (c+d x)) \left (\frac{4 (4 A+B) \sin (c) \cos (d x)}{d}+\frac{4 (4 A+B) \cos (c) \sin (d x)}{d}+\frac{4 (A+4 B) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (A+4 B) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 (8 A+13 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (8 A+13 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+2 x (13 A+8 B)+\frac{A \sin (2 c) \cos (2 d x)}{d}+\frac{A \cos (2 c) \sin (2 d x)}{d}+\frac{B}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{B}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{64 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Cos[c + d*x]^5*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(2*(13*A + 8*B)*x - (2*(8*A +
13*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(8*A + 13*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])
/d + (4*(4*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(4*A + B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*
c]*Sin[2*d*x])/d + B/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(A + 4*B)*Sin[(d*x)/2])/(d*(Cos[c/2] - S
in[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - B/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(A + 4*B)
*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(64*(B + A*Cos[c + d*x]))

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Maple [A]  time = 0.086, size = 182, normalized size = 1.1 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}Ax}{2}}+{\frac{13\,A{a}^{4}c}{2\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,B{a}^{4}x+4\,{\frac{B{a}^{4}c}{d}}+{\frac{13\,B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+4\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{B{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{4}\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)^2*(a+a*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)+13/2*a^4*A*x+13/2/d*A*a^4*c+1/d*B*a^4*sin(d*x+c)+4/d*A*a^4*sin(d*x+c)+4*B*a^
4*x+4/d*B*a^4*c+13/2/d*B*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/d*B*a^4*tan(d*x+c
)+1/d*A*a^4*tan(d*x+c)+1/2/d*B*a^4*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 1.03166, size = 269, normalized size = 1.68 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 24 \,{\left (d x + c\right )} A a^{4} + 16 \,{\left (d x + c\right )} B a^{4} - B a^{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 )} + 8 \, A a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, A a^{4} \sin \left (d x + c\right ) + 4 \, B a^{4} \sin \left (d x + c\right ) + 4 \, A a^{4} \tan \left (d x + c\right ) + 16 \, B a^{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+a*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 + 24*(d*x + c)*A*a^4 + 16*(d*x + c)*B*a^4 - B*a^4*(2*sin(d*x + c)/
(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 8*A*a^4*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 12*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 16*A*a^4*sin(d*x + c) + 4*B*a^4
*sin(d*x + c) + 4*A*a^4*tan(d*x + c) + 16*B*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.512709, size = 390, normalized size = 2.44 \begin{align*} \frac{2 \,{\left (13 \, A + 8 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{2} +{\left (8 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (8 \, A + 13 \, B\right )} a^{4} \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} + 2 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 2 \,{\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + B a^{4}\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+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(2*(13*A + 8*B)*a^4*d*x*cos(d*x + c)^2 + (8*A + 13*B)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (8*A + 13
*B)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4*cos(d*x + c)^3 + 2*(4*A + B)*a^4*cos(d*x + c)^2 + 2*(
A + 4*B)*a^4*cos(d*x + c) + B*a^4)*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+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.39989, size = 311, normalized size = 1.94 \begin{align*} \frac{{\left (13 \, A a^{4} + 8 \, B a^{4}\right )}{\left (d x + c\right )} +{\left (8 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (8 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (5 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 5 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 7 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 7 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 11 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11 \, B a^{4} \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 )^{4} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((13*A*a^4 + 8*B*a^4)*(d*x + c) + (8*A*a^4 + 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (8*A*a^4 + 13*
B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 5*B*a^4*tan(1/2*d*x + 1/2*c)^7
- 7*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 7*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*B*a^4*
tan(1/2*d*x + 1/2*c)^3 + 11*A*a^4*tan(1/2*d*x + 1/2*c) + 11*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
4 - 1)^2)/d