3.1093 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=158 \[ \frac{4 a^2 (A+3 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a^2 (7 A-15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 (A-5 C) \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}+\frac{16 a^2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt{\cos (c+d x)}} \]

[Out]

(16*a^2*A*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(7*A -
15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x
]]) + (2*(A - 5*C)*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.464138, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4114, 3044, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac{4 a^2 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (7 A-15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 (A-5 C) \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}+\frac{16 a^2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*A*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(7*A -
15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x
]]) + (2*(A - 5*C)*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*d)

Rule 4114

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sec[(e_.)
+ (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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]

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 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 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]

Rubi steps

\begin{align*} \int \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\int \frac{(a+a \cos (c+d x))^2 \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 \int \frac{(a+a \cos (c+d x))^2 \left (2 a C+\frac{1}{2} a (A-5 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{a}\\ &=\frac{2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 (A-5 C) \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{4 \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{4} a^2 (A+15 C)+\frac{1}{4} a^2 (7 A-15 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 (A-5 C) \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{4 \int \frac{\frac{1}{4} a^3 (A+15 C)+\left (\frac{1}{4} a^3 (7 A-15 C)+\frac{1}{4} a^3 (A+15 C)\right ) \cos (c+d x)+\frac{1}{4} a^3 (7 A-15 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{5 a}\\ &=\frac{2 a^2 (7 A-15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 (A-5 C) \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{8 \int \frac{\frac{5}{4} a^3 (A+3 C)+3 a^3 A \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (7 A-15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 (A-5 C) \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{1}{5} \left (8 a^2 A\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^2 (A+3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{16 a^2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^2 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (7 A-15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 (A-5 C) \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 6.4492, size = 799, normalized size = 5.06 \[ \frac{\sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \left (-\frac{(8 \cos (2 c) A+8 A-5 C+5 C \cos (2 c)) \csc (c) \sec (c)}{10 d}+\frac{C \sec (c+d x) \sin (d x) \sec (c)}{d}+\frac{2 A \cos (d x) \sin (c)}{3 d}+\frac{A \cos (2 d x) \sin (2 c)}{10 d}+\frac{2 A \cos (c) \sin (d x)}{3 d}+\frac{A \cos (2 c) \sin (2 d x)}{10 d}\right ) \cos ^{\frac{9}{2}}(c+d x)}{\cos (2 c+2 d x) A+A+2 C}-\frac{4 A \csc (c) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \left (\frac{\text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\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 ) \cos ^4(c+d x)}{5 d (\cos (2 c+2 d x) A+A+2 C)}-\frac{2 A \csc (c) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\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} \cos ^4(c+d x)}{3 d (\cos (2 c+2 d x) A+A+2 C) \sqrt{\cot ^2(c)+1}}-\frac{2 C \csc (c) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\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} \cos ^4(c+d x)}{d (\cos (2 c+2 d x) A+A+2 C) \sqrt{\cot ^2(c)+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^(9/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(-((8*A - 5*C + 8*A*Cos
[2*c] + 5*C*Cos[2*c])*Csc[c]*Sec[c])/(10*d) + (2*A*Cos[d*x]*Sin[c])/(3*d) + (A*Cos[2*d*x]*Sin[2*c])/(10*d) + (
2*A*Cos[c]*Sin[d*x])/(3*d) + (C*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (A*Cos[2*c]*Sin[2*d*x])/(10*d)))/(A + 2*C +
A*Cos[2*c + 2*d*x]) - (2*A*Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]
]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[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*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c + d*x]^4*Csc[c]*Hypergeometri
cPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c
+ d*x]^2)*Sec[d*x - ArcTan[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]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^
2]) - (4*A*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((Hypergeo
metricPFQ[{-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 + ArcT
an[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
]]))/(5*d*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [B]  time = 2.156, size = 440, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*a^2*(-12*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
32*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+15*C)*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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))-12*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)*(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*(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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(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

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Maxima [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)^(5/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} +{\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^2*cos(d*x + c)^2*sec(d*x + c)^4 + 2*C*a^2*cos(d*x + c)^2*sec(d*x + c)^3 + (A + C)*a^2*cos(d*x +
c)^2*sec(d*x + c)^2 + 2*A*a^2*cos(d*x + c)^2*sec(d*x + c) + A*a^2*cos(d*x + c)^2)*sqrt(cos(d*x + c)), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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