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

Optimal. Leaf size=245 \[ -\frac {a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac {\left (3 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}-\frac {a \left (3 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac {a^2 \left (3 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

[Out]

-a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+a*sec(d*x+c))+(3*a^2-2*b^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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)/d-a*(3*a
^2-4*b^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))*cos(d*x+c)^(1/
2)*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d+a^2*(3*a^2-5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
Pi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^3/(a+b)^2/d

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Rubi [A]  time = 0.47, antiderivative size = 245, normalized size of antiderivative = 1.00, 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, 3847, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac {a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {a \left (3 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac {\left (3 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac {a^2 \left (3 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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 3847

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 6.43, size = 319, normalized size = 1.30 \[ \frac {\frac {4 a^2 \sin (c+d x)}{b \left (b^2-a^2\right ) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))}-\frac {2 \cot (c+d x) \left (6 a^3 \sqrt {-\tan ^2(c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+2 b \left (-3 a^2+a b+2 b^2\right ) \sqrt {-\tan ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+2 b \left (3 a^2-2 b^2\right ) \sqrt {-\tan ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-3 a^2 b \sec ^{\frac {3}{2}}(c+d x)+3 a^2 b \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)-10 a b^2 \sqrt {-\tan ^2(c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+2 b^3 \sec ^{\frac {3}{2}}(c+d x)-2 b^3 \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)\right )}{b^3 (a-b) (a+b)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2*Sin[c + d*x])/(b*(-a^2 + b^2)*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]]) - (2*Cot[c + d*x]*(-3*a^2*b*Sec
[c + d*x]^(3/2) + 2*b^3*Sec[c + d*x]^(3/2) + 3*a^2*b*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) - 2*b^3*Cos[2*(c + d*
x)]*Sec[c + d*x]^(3/2) + 2*b*(3*a^2 - 2*b^2)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] +
2*b*(-3*a^2 + a*b + 2*b^2)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 6*a^3*EllipticPi
[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 10*a*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c
+ d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/((a - b)*b^3*(a + b)))/(4*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.40, size = 815, normalized size = 3.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/b^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))*a+EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)-12*a^2/b^2/(-2*a*b+2*b^2)*(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)*EllipticPi(cos(1/2*d*x+1/2*c),-2
*b/(a-b),2^(1/2))-2/b^3*a^3*(-b^2/a/(a^2-b^2)*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)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(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))-1/2*b/a/(a^2-b^
2)*(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))+1/2*b/a/(a^2-b^2)*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3
*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*
b^2)*b^3*(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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),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 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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