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

Optimal. Leaf size=147 \[ \frac{8 a^2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{7 d}-\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 \sin (c+d x)}{7 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{12 a^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

(-12*a^2*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^2*EllipticF[(c + d*x)/2, 2])/(7*d) + (2*a^2*Sin[c + d*x])/(7*
d*Cos[c + d*x]^(7/2)) + (4*a^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (8*a^2*Sin[c + d*x])/(7*d*Cos[c + d*x]
^(3/2)) + (12*a^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.172944, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4264, 3788, 3768, 3771, 2639, 4046, 2641} \[ \frac{8 a^2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}-\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 \sin (c+d x)}{7 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{12 a^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a^2*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^2*EllipticF[(c + d*x)/2, 2])/(7*d) + (2*a^2*Sin[c + d*x])/(7*
d*Cos[c + d*x]^(7/2)) + (4*a^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (8*a^2*Sin[c + d*x])/(7*d*Cos[c + d*x]
^(3/2)) + (12*a^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 4264

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]

Rule 3788

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 3768

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 3771

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 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 4046

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 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]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^2}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sec ^{\frac{5}{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{5}{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{7}{2}}(c+d x) \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{5} \left (6 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{7} \left (12 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)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{12 a^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{1}{7} \left (4 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{5} \left (6 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{12 a^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{1}{7} \left (4 a^2\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (6 a^2\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{2 a^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{12 a^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.22329, size = 531, normalized size = 3.61 \[ \frac{3 \csc (c) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{2 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{10 d}-\frac{2 \csc (c) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{7 d \sqrt{\cot ^2(c)+1}}+\cos ^{\frac{5}{2}}(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac{\sec (c) \sin (d x) \sec ^4(c+d x)}{14 d}+\frac{\sec (c) (5 \sin (c)+14 \sin (d x)) \sec ^3(c+d x)}{70 d}+\frac{\sec (c) (7 \sin (c)+10 \sin (d x)) \sec ^2(c+d x)}{35 d}+\frac{\sec (c) (10 \sin (c)+21 \sin (d x)) \sec (c+d x)}{35 d}+\frac{3 \csc (c) \sec (c)}{5 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Cos[c + d*x]^(5/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*((3*Csc[c]*Sec[c])/(5*d) + (Sec[c]*Sec[c + d*x]
^4*Sin[d*x])/(14*d) + (Sec[c]*Sec[c + d*x]^2*(7*Sin[c] + 10*Sin[d*x]))/(35*d) + (Sec[c]*Sec[c + d*x]^3*(5*Sin[
c] + 14*Sin[d*x]))/(70*d) + (Sec[c]*Sec[c + d*x]*(10*Sin[c] + 21*Sin[d*x]))/(35*d)) - (2*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 - ArcT
an[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*Sqrt[1 + Cot[c]^2]) + (3*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[C
os[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]]))/(10*d)

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Maple [B]  time = 2.522, size = 439, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*(-1/14*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+31/70*(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
))-1/224*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)^4-
1/40*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)^3-3/5*
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)-3/10*(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((a+a*sec(d*x+c))^2/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2)/cos(d*x + c)^(5/2), 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((a+a*sec(d*x+c))**2/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((a+a*sec(d*x+c))^2/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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