3.717 \(\int \frac{1}{(a+b \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=172 \[ \frac{2 \left (3 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^3 d}-\frac{2 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}-\frac{2 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}} \]

[Out]

(-2*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*d) + (2*(3*a^2 + b^2)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^3*d) - (2*a^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a
+ b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^3*(a + b)*d) + (2*Sin[c + d*x])/(3*b*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.391052, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3238, 3853, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{2 \left (3 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^3 d}-\frac{2 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}-\frac{2 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[c + d*x])*Sec[c + d*x]^(5/2)),x]

[Out]

(-2*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*d) + (2*(3*a^2 + b^2)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^3*d) - (2*a^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a
+ b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^3*(a + b)*d) + (2*Sin[c + d*x])/(3*b*d*Sqrt[Sec[c + d*x]])

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 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{1}{(a+b \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{1}{\sec ^{\frac{3}{2}}(c+d x) (b+a \sec (c+d x))} \, dx\\ &=\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}}+\frac{2 \int \frac{-\frac{3 a}{2}+\frac{1}{2} b \sec (c+d x)+\frac{1}{2} a \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{3 b}\\ &=\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}}+\frac{2 \int \frac{-\frac{3 a b}{2}-\left (-\frac{3 a^2}{2}-\frac{b^2}{2}\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 b^3}-\frac{a^3 \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{b^3}\\ &=\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}}-\frac{a \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{b^2}+\frac{\left (3 a^2+b^2\right ) \int \sqrt{\sec (c+d x)} \, dx}{3 b^3}-\frac{\left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^3}\\ &=-\frac{2 a^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^3 (a+b) d}+\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}}-\frac{\left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{b^2}+\frac{\left (\left (3 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^3}\\ &=-\frac{2 a \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^2 d}+\frac{2 \left (3 a^2+b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 b^3 d}-\frac{2 a^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^3 (a+b) d}+\frac{2 \sin (c+d x)}{3 b d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.12514, size = 198, normalized size = 1.15 \[ -\frac{\cot (c+d x) \left (12 a^2 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+6 a b \sec ^{\frac{3}{2}}(c+d x)-6 a b \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)+4 b (3 a-b) \sqrt{-\tan ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-12 a b \sqrt{-\tan ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-b^2 \sqrt{\sec (c+d x)}+b^2 \cos (3 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)\right )}{6 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Cos[c + d*x])*Sec[c + d*x]^(5/2)),x]

[Out]

-(Cot[c + d*x]*(-(b^2*Sqrt[Sec[c + d*x]]) + 6*a*b*Sec[c + d*x]^(3/2) - 6*a*b*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/
2) + b^2*Cos[3*(c + d*x)]*Sec[c + d*x]^(3/2) - 12*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c +
d*x]^2] + 4*(3*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 12*a^2*EllipticPi[-(
a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(6*b^3*d)

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Maple [B]  time = 3.504, size = 516, 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(1/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2),x)

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((4*a*b^2-4*b^3)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^4+(-2*a*b^2+2*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+3*(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))*a^3-3*a^2*b*(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))+(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))*a*b^2-b^3*(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))+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/
2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/
2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/b^3/(a-b)/(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*cos(d*x + c) + a)*sec(d*x + c)^(5/2)), x)

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Fricas [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(1/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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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(1/(a+b*cos(d*x+c))/sec(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{1}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*cos(d*x + c) + a)*sec(d*x + c)^(5/2)), x)