3.141 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=164 \[ \frac {a^{5/2} (25 A+38 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[Out]

1/8*a^(5/2)*(25*A+38*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/3*a*A*cos(d*x+c)^2*(a+a*sec(d*x+
c))^(3/2)*sin(d*x+c)/d+1/24*a^3*(49*A+54*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a^2*(3*A+2*B)*cos(d*x+c)*s
in(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.46, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4017, 4015, 3774, 203} \[ \frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}+\frac {a^{5/2} (25 A+38 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a^(5/2)*(25*A + 38*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(49*A + 54*B)*Sin
[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(3*A + 2*B)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*
x])/(4*d) + (a*A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 4015

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

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]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {3}{2} a (3 A+2 B)+\frac {1}{2} a (A+6 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (49 A+54 B)+\frac {1}{4} a^2 (13 A+30 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{16} \left (a^2 (25 A+38 B)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {\left (a^3 (25 A+38 B)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}\\ &=\frac {a^{5/2} (25 A+38 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C]  time = 1.04, size = 312, normalized size = 1.90 \[ -\frac {a^2 \cos (c+d x) \sqrt {a (\sec (c+d x)+1)} \left (-192 A \tan (c+d x) \sqrt {1-\sec (c+d x)} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};1-\sec (c+d x)\right )-165 A \sin (c+d x) \sqrt {1-\sec (c+d x)}-31 A \sin (2 (c+d x)) \sqrt {1-\sec (c+d x)}+8 A \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec (c+d x)}-165 A \tan (c+d x) \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )-576 B \tan (c+d x) \sqrt {1-\sec (c+d x)} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-\sec (c+d x)\right )+18 B \sin (c+d x) \sqrt {1-\sec (c+d x)}+54 B \sin (2 (c+d x)) \sqrt {1-\sec (c+d x)}-126 B \tan (c+d x) \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )\right )}{72 d (\cos (c+d x)+1) \sqrt {1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

-1/72*(a^2*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(-165*A*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 18*B*Sqrt[1 -
Sec[c + d*x]]*Sin[c + d*x] + 8*A*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] - 31*A*Sqrt[1 - Sec[c + d
*x]]*Sin[2*(c + d*x)] + 54*B*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)] - 165*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*T
an[c + d*x] - 126*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 576*B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Se
c[c + d*x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x] - 192*A*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Sqrt[
1 - Sec[c + d*x]]*Tan[c + d*x]))/(d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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fricas [A]  time = 0.52, size = 380, normalized size = 2.32 \[ \left [\frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/48*(3*((25*A + 38*B)*a^2*cos(d*x + c) + (25*A + 38*B)*a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*
(8*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 22*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(3*((25*A + 38*B)*a^2*cos(d*x + c) + (25*A
+ 38*B)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (8
*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 22*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x
+ c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

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giac [B]  time = 2.48, size = 905, normalized size = 5.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/48*(3*(25*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 38*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 3*(25*A*sqrt(-a)*a^2*sgn(cos(d*
x + c)) + 38*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(75*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a))^10*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 114*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 1125*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 1710*sqrt(2)*(sqrt(-a)*tan(1/2*d*x +
1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 6174*sqrt(2)*(sqrt(-a)*tan(
1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 6804*sqrt(2)*(sqr
t(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 4314*sq
rt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^6*sgn(cos(d*x + c))
- 4284*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^6*sgn(cos
(d*x + c)) + 807*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^
7*sgn(cos(d*x + c)) + 858*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sq
rt(-a)*a^7*sgn(cos(d*x + c)) - 49*sqrt(2)*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 54*sqrt(2)*B*sqrt(-a)*a^8*sgn(cos
(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x
+ 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d

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maple [B]  time = 1.71, size = 583, normalized size = 3.55 \[ -\frac {\left (75 A \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+114 B \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+150 A \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \cos \left (d x +c \right )+228 B \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \cos \left (d x +c \right )+75 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+114 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+64 A \left (\cos ^{6}\left (d x +c \right )\right )+208 A \left (\cos ^{5}\left (d x +c \right )\right )+96 B \left (\cos ^{5}\left (d x +c \right )\right )+328 A \left (\cos ^{4}\left (d x +c \right )\right )+432 B \left (\cos ^{4}\left (d x +c \right )\right )-600 A \left (\cos ^{3}\left (d x +c \right )\right )-528 B \left (\cos ^{3}\left (d x +c \right )\right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{192 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

-1/192/d*(75*A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*cos(d*x+c)^2+114*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*ar
ctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*cos(d*x+c)^2+150*A*sin(d
*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x
+c)*2^(1/2))*2^(1/2)*cos(d*x+c)+228*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*cos(d*x+c)+75*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(5/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+114*
B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos
(d*x+c)*2^(1/2))*sin(d*x+c)+64*A*cos(d*x+c)^6+208*A*cos(d*x+c)^5+96*B*cos(d*x+c)^5+328*A*cos(d*x+c)^4+432*B*co
s(d*x+c)^4-600*A*cos(d*x+c)^3-528*B*cos(d*x+c)^3)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)*
a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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