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

Optimal. Leaf size=220 \[ \frac{(A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}-\frac{(A+9 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-6 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

((A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((A + 3*B)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(
5*d*(a + a*Sec[c + d*x])^3) + ((A - 6*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((
A + 9*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.48983, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4019, 3787, 3771, 2639, 2641} \[ -\frac{(A+9 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-6 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((A + 3*B)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(
5*d*(a + a*Sec[c + d*x])^3) + ((A - 6*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((
A + 9*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + 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] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rule 3787

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

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 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{\sec ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3}{2} a (A-B)+\frac{1}{2} a (A+9 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-6 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} a^2 (A-6 B)+\frac{1}{2} a^2 (4 A+21 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-6 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A+9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{\frac{3}{4} a^3 (A+9 B)+\frac{5}{4} a^3 (A+3 B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^6}\\ &=\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-6 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A+9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(A+3 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}+\frac{(A+9 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-6 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A+9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\left ((A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}+\frac{\left ((A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{(A+9 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}+\frac{(A+3 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}+\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-6 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A+9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.88495, size = 919, normalized size = 4.18 \[ -\frac{\sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}-\frac{3 \sqrt{2} B e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac{2 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac{2 B \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac{\sec ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)) \left (\frac{2 \sec \left (\frac{c}{2}\right ) \left (B \sin \left (\frac{d x}{2}\right )-A \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}+\frac{2 (B-A) \tan \left (\frac{c}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \left (2 A \sin \left (\frac{d x}{2}\right )+3 B \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d}+\frac{4 (2 A+3 B) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+3 B \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{2 (A+9 B) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{5 d}+\frac{4 (A+3 B) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6
*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(15*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a
+ a*Sec[c + d*x])^3) - (3*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*
x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hyperg
eometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(5*d*E^(I*d*x)
*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (2*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*Elliptic
F[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Se
c[c + d*x])^3) + (2*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[
c + d*x]^(5/2)*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)
/2]^6*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*((-2*(A + 9*B)*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) + (2*Sec[c/2]*S
ec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/
2] + 3*B*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(2*A*Sin[(d*x)/2] + 3*B*Sin[(d*x)/2]))/(15*d)
+ (4*(A + 3*B)*Tan[c/2])/(3*d) + (4*(2*A + 3*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(-A + B)*Sec[c/2 +
(d*x)/2]^4*Tan[c/2])/(5*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3)

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Maple [A]  time = 2.099, size = 451, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2*d*x+1/2*c)^8-10*A*cos(1/2*d*x+1/2*c
)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*A*c
os(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))+108*B*cos(1/2*d*x+1/2*c)^8-30*B*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+54*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-22*A*cos(1/2*d*x+1/2*c)^6-138*B*cos(1/
2*d*x+1/2*c)^6+6*A*cos(1/2*d*x+1/2*c)^4+24*B*cos(1/2*d*x+1/2*c)^4+7*A*cos(1/2*d*x+1/2*c)^2+3*B*cos(1/2*d*x+1/2
*c)^2-3*A+3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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