3.77 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=165 \[ \frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a^4 (A+4 B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(8 A-3 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{(2 A+B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{1}{2} a^4 x (12 A+13 B)+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.409852, antiderivative size = 165, 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} \[ \frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a^4 (A+4 B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(8 A-3 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{(2 A+B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{1}{2} a^4 x (12 A+13 B)+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(a^4*(12*A + 13*B)*x)/2 + (a^4*(A + 4*B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(2*A + B)*Sin[c + d*x])/(2*d) + (a*
A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((2*A + B)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])
^2*Sin[c + d*x])/(2*d) - ((8*A - 3*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(6*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 ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (3 a (2 A+B)-a (A-3 B) \sec (c+d x)) \, dx\\ &=\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{1}{6} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (11 A+9 B)-a^2 (8 A-3 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (2 A+B)+6 a^3 (A+4 B) \sec (c+d x)\right ) \, dx\\ &=\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac{1}{6} \int \left (-3 a^4 (12 A+13 B)-6 a^4 (A+4 B) \sec (c+d x)\right ) \, dx\\ &=\frac{1}{2} a^4 (12 A+13 B) x+\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^4 (A+4 B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (12 A+13 B) x+\frac{a^4 (A+4 B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 1.85038, size = 342, normalized size = 2.07 \[ \frac{a^4 \cos ^5(c+d x) \sec ^8\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 (A+B \sec (c+d x)) \left (\frac{3 (27 A+16 B) \sin (c) \cos (d x)}{d}+\frac{3 (4 A+B) \sin (2 c) \cos (2 d x)}{d}+\frac{3 (27 A+16 B) \cos (c) \sin (d x)}{d}+\frac{3 (4 A+B) \cos (2 c) \sin (2 d x)}{d}-\frac{12 (A+4 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{12 (A+4 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{A \sin (3 c) \cos (3 d x)}{d}+\frac{A \cos (3 c) \sin (3 d x)}{d}+72 A x+\frac{12 B \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{12 B \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+78 B x\right )}{192 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(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])*(72*A*x + 78*B*x - (12*(A + 4
*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (12*(A + 4*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d +
(3*(27*A + 16*B)*Cos[d*x]*Sin[c])/d + (3*(4*A + B)*Cos[2*d*x]*Sin[2*c])/d + (A*Cos[3*d*x]*Sin[3*c])/d + (3*(27
*A + 16*B)*Cos[c]*Sin[d*x])/d + (3*(4*A + B)*Cos[2*c]*Sin[2*d*x])/d + (A*Cos[3*c]*Sin[3*d*x])/d + (12*B*Sin[(d
*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (12*B*Sin[(d*x)/2])/(d*(Cos[c/2] + S
in[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(192*(B + A*Cos[c + d*x]))

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Maple [A]  time = 0.081, size = 190, normalized size = 1.2 \begin{align*}{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{20\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{13\,B{a}^{4}x}{2}}+{\frac{13\,B{a}^{4}c}{2\,d}}+2\,{\frac{A{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+6\,{a}^{4}Ax+6\,{\frac{A{a}^{4}c}{d}}+4\,{\frac{B{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{4}\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

1/3/d*A*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*A*a^4*sin(d*x+c)+1/2/d*B*a^4*sin(d*x+c)*cos(d*x+c)+13/2*B*a^4*x+13/
2/d*B*a^4*c+2/d*A*a^4*sin(d*x+c)*cos(d*x+c)+6*a^4*A*x+6/d*A*a^4*c+4/d*B*a^4*sin(d*x+c)+4/d*B*a^4*ln(sec(d*x+c)
+tan(d*x+c))+1/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/d*B*a^4*tan(d*x+c)

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Maxima [A]  time = 1.04287, size = 252, normalized size = 1.53 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 48 \,{\left (d x + c\right )} A a^{4} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 72 \,{\left (d x + c\right )} B a^{4} - 6 \, 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 )} - 24 \, 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 )} - 72 \, A a^{4} \sin \left (d x + c\right ) - 48 \, B a^{4} \sin \left (d x + c\right ) - 12 \, B a^{4} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 48*(d*x + c)*A*
a^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 72*(d*x + c)*B*a^4 - 6*A*a^4*(log(sin(d*x + c) + 1) - log(sin
(d*x + c) - 1)) - 24*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 72*A*a^4*sin(d*x + c) - 48*B*a^4*
sin(d*x + c) - 12*B*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.514288, size = 383, normalized size = 2.32 \begin{align*} \frac{3 \,{\left (12 \, A + 13 \, B\right )} a^{4} d x \cos \left (d x + c\right ) + 3 \,{\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, A a^{4} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \,{\left (5 \, A + 3 \, B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, B a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*(12*A + 13*B)*a^4*d*x*cos(d*x + c) + 3*(A + 4*B)*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 3*(A + 4*B)*a
^4*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*A*a^4*cos(d*x + c)^3 + 3*(4*A + B)*a^4*cos(d*x + c)^2 + 8*(5*A + 3
*B)*a^4*cos(d*x + c) + 6*B*a^4)*sin(d*x + c))/(d*cos(d*x + c))

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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)**3*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30382, size = 305, normalized size = 1.85 \begin{align*} -\frac{\frac{12 \, B a^{4} \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} - 3 \,{\left (12 \, A a^{4} + 13 \, B a^{4}\right )}{\left (d x + c\right )} - 6 \,{\left (A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 6 \,{\left (A a^{4} + 4 \, 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 (30 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 76 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 48 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, 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 )^{2} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(12*B*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(12*A*a^4 + 13*B*a^4)*(d*x + c) - 6*(A*a^
4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
30*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^4
*tan(1/2*d*x + 1/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)
^2 + 1)^3)/d