3.4.65 \(\int \frac {(a+a \sec (c+d x))^2}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [365]

Optimal. Leaf size=121 \[ -\frac {16 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 a^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-16/5*a^2*(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+4/3*a^2*(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^2*sin(d*x+c)/d/cos(d
*x+c)^(5/2)+4/3*a^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+16/5*a^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4349, 3873, 3853, 3856, 2720, 4131, 2719} \begin {gather*} \frac {4 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {16 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16 a^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

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

Rule 2720

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

Rule 3853

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.19, size = 503, normalized size = 4.16 \begin {gather*} \cos ^{\frac {5}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (\frac {4 \csc (c) \sec (c)}{5 d}+\frac {\sec (c) \sec ^3(c+d x) \sin (d x)}{10 d}+\frac {\sec (c) \sec ^2(c+d x) (3 \sin (c)+10 \sin (d x))}{30 d}+\frac {\sec (c) \sec (c+d x) (5 \sin (c)+12 \sin (d x))}{15 d}\right )-\frac {\cos ^2(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d \sqrt {1+\cot ^2(c)}}+\frac {2 \cos ^2(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Cos[c + d*x]^(5/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*((4*Csc[c]*Sec[c])/(5*d) + (Sec[c]*Sec[c + d*x]
^3*Sin[d*x])/(10*d) + (Sec[c]*Sec[c + d*x]^2*(3*Sin[c] + 10*Sin[d*x]))/(30*d) + (Sec[c]*Sec[c + d*x]*(5*Sin[c]
+ 12*Sin[d*x]))/(15*d)) - (Cos[c + d*x]^2*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*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*Sqrt[
1 + Cot[c]^2]) + (2*Cos[c + d*x]^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^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]]]*Sq
rt[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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(157)=314\).
time = 0.13, size = 386, normalized size = 3.19

method	result	size
default	\(-\frac {8 \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^{2} \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{12 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {17 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}{30 \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 )}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{80 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}-\frac {4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \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 )}}-\frac {2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \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 )}}\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}\)	\(386\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-1/12*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^2+17/30*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))-1/80*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^3-4
/5*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-2/5*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)
*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.80, size = 202, normalized size = 1.67 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (24 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*I*sqrt(2)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*
a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*I*sqrt(2)*a^2*cos(d*x + c)^3
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*I*sqrt(2)*a^2*cos(d*x
+ c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (24*a^2*cos(d*x + c
)^2 + 10*a^2*cos(d*x + c) + 3*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {2 \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {1}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(2*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(sec(c + d*x)**2/cos(c + d*x)**(3/2), x) + Int
egral(cos(c + d*x)**(-3/2), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)

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Mupad [B]
time = 1.38, size = 114, normalized size = 0.94 \begin {gather*} \frac {6\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*a^2*cos(c + d*x)*sin(c + d*x)*hypergeom(
[-3/4, 1/2], 1/4, cos(c + d*x)^2) + 30*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x
)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2))

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