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

Optimal. Leaf size=101 \[ \frac {2 a (A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d} \]

[Out]

2/5*a*(3*A+5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*
(A+B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/5*a*A*cos(d*x+
c)^(3/2)*sin(d*x+c)/d+2/3*a*(A+B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2954, 2968, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 a (A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(3*A + 5*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(A + B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*(A +
B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx &=\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (B+A \cos (c+d x)) \, dx\\ &=\int \sqrt {\cos (c+d x)} \left (a B+(a A+a B) \cos (c+d x)+a A \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\cos (c+d x)} \left (\frac {1}{2} a (3 A+5 B)+\frac {5}{2} a (A+B) \cos (c+d x)\right ) \, dx\\ &=\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+(a (A+B)) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{5} (a (3 A+5 B)) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} (a (A+B)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.22, size = 830, normalized size = 8.22 \[ a \left (\sqrt {\cos (c+d x)} (\cos (c+d x)+1) \left (-\frac {(3 A+5 B) \cot (c)}{5 d}+\frac {(A+B) \cos (d x) \sin (c)}{3 d}+\frac {A \cos (2 d x) \sin (2 c)}{10 d}+\frac {(A+B) \cos (c) \sin (d x)}{3 d}+\frac {A \cos (2 c) \sin (2 d x)}{10 d}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-\frac {3 A (\cos (c+d x)+1) \csc (c) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt {\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}} \sqrt {\tan ^2(c)+1}}-\frac {\frac {2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac {\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {\tan ^2(c)+1}}}{\sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{10 d}-\frac {B (\cos (c+d x)+1) \csc (c) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt {\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}} \sqrt {\tan ^2(c)+1}}-\frac {\frac {2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac {\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {\tan ^2(c)+1}}}{\sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}-\frac {A (\cos (c+d x)+1) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {-\sqrt {\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \sqrt {\cot ^2(c)+1}}-\frac {B (\cos (c+d x)+1) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {-\sqrt {\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \sqrt {\cot ^2(c)+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(-1/5*((3*A + 5*B)*Cot[c])/d + ((A + B)*Cos[d*x]
*Sin[c])/(3*d) + (A*Cos[2*d*x]*Sin[2*c])/(10*d) + ((A + B)*Cos[c]*Sin[d*x])/(3*d) + (A*Cos[2*c]*Sin[2*d*x])/(1
0*d)) - (A*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
+ (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*S
in[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (B*(1 + Cos[c + d*x
])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - Arc
Tan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*
Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (3*A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)
/2]^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/
(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]
*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^
2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*S
qrt[1 + Tan[c]^2]]))/(10*d) - (B*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2, -1/
4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]
*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]
^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 +
Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d))

________________________________________________________________________________________

fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B a \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + {\left (A + B\right )} a \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A a \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*a*cos(d*x + c)^2*sec(d*x + c)^2 + (A + B)*a*cos(d*x + c)^2*sec(d*x + c) + A*a*cos(d*x + c)^2)*sqrt
(cos(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 4.07, size = 355, normalized size = 3.51 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (44 A +20 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-16 A -10 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(-24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(44*A+20*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-16*A-10*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*A*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+5*B*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.36, size = 128, normalized size = 1.27 \[ \frac {2\,A\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*B*a*(cos(c + d*x)^(1/2)*sin
(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*B*a*ellipticE(c/2 + (d*x)/2, 2))/d - (2*A*a*cos(c + d*x)^
(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________