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

Optimal. Leaf size=151 \[ -\frac{5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac{a^4 (13 A+12 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+a^4 x (4 A+B)+\frac{a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[Out]

a^4*(4*A + B)*x + (a^4*(13*A + 12*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A + 2*B)*Sin[c + d*x])/(2*d) + (a*
B*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((A + 2*B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((9
*A + 11*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.367918, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4018, 3996, 3770} \[ -\frac{5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac{a^4 (13 A+12 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+a^4 x (4 A+B)+\frac{a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*(4*A + B)*x + (a^4*(13*A + 12*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A + 2*B)*Sin[c + d*x])/(2*d) + (a*
B*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((A + 2*B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((9
*A + 11*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(3*d)

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

Mathematica [B]  time = 6.44965, size = 1202, normalized size = 7.96 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A + B)*x*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(16*(B + A*Cos[c
+ d*x])) + ((-13*A - 12*B)*Cos[c + d*x]^5*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(
a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(32*d*(B + A*Cos[c + d*x])) + ((13*A + 12*B)*Cos[c + d*x]^5*Log[Co
s[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(32*d
*(B + A*Cos[c + d*x])) + (A*Cos[d*x]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c +
d*x])*Sin[c])/(16*d*(B + A*Cos[c + d*x])) + (A*Cos[c]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x]
)^4*(A + B*Sec[c + d*x])*Sin[d*x])/(16*d*(B + A*Cos[c + d*x])) + (B*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a
*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(96*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2
+ (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c
+ d*x])*(3*A*Cos[c/2] + 13*B*Cos[c/2] - 3*A*Sin[c/2] - 11*B*Sin[c/2]))/(192*d*(B + A*Cos[c + d*x])*(Cos[c/2]
- Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c +
d*x])^4*(A + B*Sec[c + d*x])*(3*A*Sin[(d*x)/2] + 5*B*Sin[(d*x)/2]))/(12*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Si
n[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (B*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x
])^4*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(96*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] +
Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(-3
*A*Cos[c/2] - 13*B*Cos[c/2] - 3*A*Sin[c/2] - 11*B*Sin[c/2]))/(192*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])
*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A
+ B*Sec[c + d*x])*(3*A*Sin[(d*x)/2] + 5*B*Sin[(d*x)/2]))/(12*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])*(Co
s[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.084, size = 189, normalized size = 1.3 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+B{a}^{4}x+{\frac{B{a}^{4}c}{d}}+4\,{a}^{4}Ax+4\,{\frac{A{a}^{4}c}{d}}+6\,{\frac{B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{13\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{20\,B{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+4\,{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,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))^4*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.01326, size = 317, normalized size = 2.1 \begin{align*} \frac{48 \,{\left (d x + c\right )} A a^{4} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 12 \,{\left (d x + c\right )} B a^{4} - 3 \, A 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 )} - 12 \, 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 )} + 36 \, 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 )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 48 \, A a^{4} \tan \left (d x + c\right ) + 72 \, 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)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*A*a^4 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 12*(d*x + c)*B*a^4 - 3*A*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)) - 12*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)) + 36*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)) + 12*A*a^4*sin(d*x + c) + 48*A*a^4*tan(
d*x + c) + 72*B*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.509542, size = 405, normalized size = 2.68 \begin{align*} \frac{12 \,{\left (4 \, A + B\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (13 \, A + 12 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (13 \, A + 12 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 8 \,{\left (3 \, A + 5 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \,{\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, B a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.34172, size = 306, normalized size = 2.03 \begin{align*} \frac{\frac{12 \, A 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} + 6 \,{\left (4 \, A a^{4} + B a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (13 \, A a^{4} + 12 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (13 \, A a^{4} + 12 \, 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 (21 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 48 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 76 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 54 \, 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)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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