3.6 \(\int \frac{A+B \sec (c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d}+\frac{6 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}} \]

[Out]

(6*A*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*B*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^3*d) + (2*A*Sin[c + d*x])/(5*b*d*(b*Sec[c + d*x])^(3/2)) +
(2*B*Sin[c + d*x])/(3*b^2*d*Sqrt[b*Sec[c + d*x]])

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Rubi [A]  time = 0.10415, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3787, 3769, 3771, 2639, 2641} \[ \frac{6 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{2 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*A*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*B*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^3*d) + (2*A*Sin[c + d*x])/(5*b*d*(b*Sec[c + d*x])^(3/2)) +
(2*B*Sin[c + d*x])/(3*b^2*d*Sqrt[b*Sec[c + d*x]])

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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 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+B \sec (c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=A \int \frac{1}{(b \sec (c+d x))^{5/2}} \, dx+\frac{B \int \frac{1}{(b \sec (c+d x))^{3/2}} \, dx}{b}\\ &=\frac{2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{(3 A) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{5 b^2}+\frac{B \int \sqrt{b \sec (c+d x)} \, dx}{3 b^3}\\ &=\frac{2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{(3 A) \int \sqrt{\cos (c+d x)} \, dx}{5 b^2 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{\left (B \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^3}\\ &=\frac{6 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d}+\frac{2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.508454, size = 88, normalized size = 0.6 \[ \frac{2 \left (5 B \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} (3 A \cos (c+d x)+5 B)+9 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d \cos ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.198, size = 482, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(9*I*A*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*sin(d*x+c)-9*I*A*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d
*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+9*I*A*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-9*I*A*sin(d*x+c)*(1/(cos(d*x+c)+1))^(
1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*B*EllipticF(I*(-1+cos(d*x
+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*A*cos(d*x+c)^4-5*B*
cos(d*x+c)^3-6*A*cos(d*x+c)^2+9*A*cos(d*x+c)+5*B*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^3/(b/cos(d*x+c))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)/(b*sec(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{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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