3.2.69 \(\int \frac {1}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [169]

Optimal. Leaf size=223 \[ \frac {b \cos (e+f x) \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 b (2 a+b) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {2 (2 a+b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 a^2 (a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 a (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

1/3*b*cos(f*x+e)*sin(f*x+e)/a/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)+2/3*b*(2*a+b)*cos(f*x+e)*sin(f*x+e)/a^2/(a+b)^2
/f/(a+b*sin(f*x+e)^2)^(1/2)+2/3*(2*a+b)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*(a+
b*sin(f*x+e)^2)^(1/2)/a^2/(a+b)^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)-1/3*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticF(s
in(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/a/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3263, 3252, 3251, 3257, 3256, 3262, 3261} \begin {gather*} \frac {2 b (2 a+b) \sin (e+f x) \cos (e+f x)}{3 a^2 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {2 (2 a+b) \sqrt {a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{3 a^2 f (a+b)^2 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {b \sin (e+f x) \cos (e+f x)}{3 a f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{3 a f (a+b) \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^(-5/2),x]

[Out]

(b*Cos[e + f*x]*Sin[e + f*x])/(3*a*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*b*(2*a + b)*Cos[e + f*x]*Sin[e
+ f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + (2*(2*a + b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*S
in[e + f*x]^2])/(3*a^2*(a + b)^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b*S
in[e + f*x]^2)/a])/(3*a*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3251

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3252

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Dist[
1/(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b
- a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 3256

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]/f)*EllipticE[e + f*x, -b/a], x] /
; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3257

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + b*(Sin
[e + f*x]^2/a)], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3261

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3262

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3263

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^
(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && N
eQ[a + b, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.93, size = 172, normalized size = 0.77 \begin {gather*} \frac {2 a^2 (2 a+b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} E\left (e+f x\left |-\frac {b}{a}\right .\right )-a^2 (a+b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} F\left (e+f x\left |-\frac {b}{a}\right .\right )-\sqrt {2} b \left (-5 a^2-5 a b-b^2+b (2 a+b) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{3 a^2 (a+b)^2 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x]^2)^(-5/2),x]

[Out]

(2*a^2*(2*a + b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] - a^2*(a + b)*((2*a + b -
b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(-5*a^2 - 5*a*b - b^2 + b*(2*a + b)*Cos[2
*(e + f*x)])*Sin[2*(e + f*x)])/(3*a^2*(a + b)^2*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(546\) vs. \(2(245)=490\).
time = 9.72, size = 547, normalized size = 2.45

method	result	size
default	\(-\frac {\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b \left (\sin ^{2}\left (f x +e \right )\right )+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2} \left (\sin ^{2}\left (f x +e \right )\right )-4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b \left (\sin ^{2}\left (f x +e \right )\right )-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2} \left (\sin ^{2}\left (f x +e \right )\right )+4 a \,b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+2 b^{3} \left (\sin ^{5}\left (f x +e \right )\right )+\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, a^{3}+a^{2} \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +5 a^{2} b \left (\sin ^{3}\left (f x +e \right )\right )-a \,b^{2} \left (\sin ^{3}\left (f x +e \right )\right )-2 b^{3} \left (\sin ^{3}\left (f x +e \right )\right )-5 a^{2} b \sin \left (f x +e \right )-3 a \,b^{2} \sin \left (f x +e \right )}{3 \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} a^{2} \left (a +b \right )^{2} \cos \left (f x +e \right ) f}\)	\(547\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*((cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b*sin(f*x+e)^
2+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2*sin(f*x+e)^2-4*
(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b*sin(f*x+e)^2-2*(c
os(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2*sin(f*x+e)^2+4*a*b^
2*sin(f*x+e)^5+2*b^3*sin(f*x+e)^5+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a
*b)^(1/2))*a^3+a^2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-4*
(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3-2*(cos(f*x+e)^2)^(1
/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b+5*a^2*b*sin(f*x+e)^3-a*b^2*sin(f*x
+e)^3-2*b^3*sin(f*x+e)^3-5*a^2*b*sin(f*x+e)-3*a*b^2*sin(f*x+e))/(a+b*sin(f*x+e)^2)^(3/2)/a^2/(a+b)^2/cos(f*x+e
)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(-5/2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.22, size = 1531, normalized size = 6.87 \begin {gather*} \frac {{\left (2 \, {\left (2 i \, a^{3} b^{2} + 5 i \, a^{2} b^{3} + 4 i \, a b^{4} + i \, b^{5} + {\left (2 i \, a b^{4} + i \, b^{5}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (2 i \, a^{2} b^{3} + 3 i \, a b^{4} + i \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left (-4 i \, a^{4} b - 12 i \, a^{3} b^{2} - 13 i \, a^{2} b^{3} - 6 i \, a b^{4} - i \, b^{5} + {\left (-4 i \, a^{2} b^{3} - 4 i \, a b^{4} - i \, b^{5}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (4 i \, a^{3} b^{2} + 8 i \, a^{2} b^{3} + 5 i \, a b^{4} + i \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left (-2 i \, a^{3} b^{2} - 5 i \, a^{2} b^{3} - 4 i \, a b^{4} - i \, b^{5} + {\left (-2 i \, a b^{4} - i \, b^{5}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (-2 i \, a^{2} b^{3} - 3 i \, a b^{4} - i \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left (4 i \, a^{4} b + 12 i \, a^{3} b^{2} + 13 i \, a^{2} b^{3} + 6 i \, a b^{4} + i \, b^{5} + {\left (4 i \, a^{2} b^{3} + 4 i \, a b^{4} + i \, b^{5}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (-4 i \, a^{3} b^{2} - 8 i \, a^{2} b^{3} - 5 i \, a b^{4} - i \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left (-3 i \, a^{4} b - 11 i \, a^{3} b^{2} - 15 i \, a^{2} b^{3} - 9 i \, a b^{4} - 2 i \, b^{5} + {\left (-3 i \, a^{2} b^{3} - 5 i \, a b^{4} - 2 i \, b^{5}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (-3 i \, a^{3} b^{2} - 8 i \, a^{2} b^{3} - 7 i \, a b^{4} - 2 i \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left (-6 i \, a^{5} - 17 i \, a^{4} b - 17 i \, a^{3} b^{2} - 7 i \, a^{2} b^{3} - i \, a b^{4} + {\left (-6 i \, a^{3} b^{2} - 5 i \, a^{2} b^{3} - i \, a b^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (6 i \, a^{4} b + 11 i \, a^{3} b^{2} + 6 i \, a^{2} b^{3} + i \, a b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left (3 i \, a^{4} b + 11 i \, a^{3} b^{2} + 15 i \, a^{2} b^{3} + 9 i \, a b^{4} + 2 i \, b^{5} + {\left (3 i \, a^{2} b^{3} + 5 i \, a b^{4} + 2 i \, b^{5}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 i \, a^{3} b^{2} + 8 i \, a^{2} b^{3} + 7 i \, a b^{4} + 2 i \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left (6 i \, a^{5} + 17 i \, a^{4} b + 17 i \, a^{3} b^{2} + 7 i \, a^{2} b^{3} + i \, a b^{4} + {\left (6 i \, a^{3} b^{2} + 5 i \, a^{2} b^{3} + i \, a b^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (-6 i \, a^{4} b - 11 i \, a^{3} b^{2} - 6 i \, a^{2} b^{3} - i \, a b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - {\left (2 \, {\left (2 \, a b^{4} + b^{5}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} b^{3} + 7 \, a b^{4} + 2 \, b^{5}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{4} b^{4} + 2 \, a^{3} b^{5} + a^{2} b^{6}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{5} b^{3} + 3 \, a^{4} b^{4} + 3 \, a^{3} b^{5} + a^{2} b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b^{2} + 4 \, a^{5} b^{3} + 6 \, a^{4} b^{4} + 4 \, a^{3} b^{5} + a^{2} b^{6}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/3*((2*(2*I*a^3*b^2 + 5*I*a^2*b^3 + 4*I*a*b^4 + I*b^5 + (2*I*a*b^4 + I*b^5)*cos(f*x + e)^4 - 2*(2*I*a^2*b^3 +
3*I*a*b^4 + I*b^5)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (-4*I*a^4*b - 12*I*a^3*b^2 - 13*I*a^2*b^3
- 6*I*a*b^4 - I*b^5 + (-4*I*a^2*b^3 - 4*I*a*b^4 - I*b^5)*cos(f*x + e)^4 + 2*(4*I*a^3*b^2 + 8*I*a^2*b^3 + 5*I*
a*b^4 + I*b^5)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt(
(2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b
^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*(-2*I*a^3*b^2 - 5*I*a^2*b^3 - 4*I*a*b^4 - I*b^5 + (-2*I*a*b^4 - I*b^5)*co
s(f*x + e)^4 - 2*(-2*I*a^2*b^3 - 3*I*a*b^4 - I*b^5)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (4*I*a^4*
b + 12*I*a^3*b^2 + 13*I*a^2*b^3 + 6*I*a*b^4 + I*b^5 + (4*I*a^2*b^3 + 4*I*a*b^4 + I*b^5)*cos(f*x + e)^4 + 2*(-4
*I*a^3*b^2 - 8*I*a^2*b^3 - 5*I*a*b^4 - I*b^5)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a
+ b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a
^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*(-3*I*a^4*b - 11*I*a^3*b^2 - 15*I*a^2*b^3
- 9*I*a*b^4 - 2*I*b^5 + (-3*I*a^2*b^3 - 5*I*a*b^4 - 2*I*b^5)*cos(f*x + e)^4 - 2*(-3*I*a^3*b^2 - 8*I*a^2*b^3 -
7*I*a*b^4 - 2*I*b^5)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (-6*I*a^5 - 17*I*a^4*b - 17*I*a^3*b^2 -
7*I*a^2*b^3 - I*a*b^4 + (-6*I*a^3*b^2 - 5*I*a^2*b^3 - I*a*b^4)*cos(f*x + e)^4 + 2*(6*I*a^4*b + 11*I*a^3*b^2 +
6*I*a^2*b^3 + I*a*b^4)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcs
in(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2
*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*(3*I*a^4*b + 11*I*a^3*b^2 + 15*I*a^2*b^3 + 9*I*a*b^4 + 2*I*b^5 +
(3*I*a^2*b^3 + 5*I*a*b^4 + 2*I*b^5)*cos(f*x + e)^4 - 2*(3*I*a^3*b^2 + 8*I*a^2*b^3 + 7*I*a*b^4 + 2*I*b^5)*cos(f
*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (6*I*a^5 + 17*I*a^4*b + 17*I*a^3*b^2 + 7*I*a^2*b^3 + I*a*b^4 + (6*
I*a^3*b^2 + 5*I*a^2*b^3 + I*a*b^4)*cos(f*x + e)^4 + 2*(-6*I*a^4*b - 11*I*a^3*b^2 - 6*I*a^2*b^3 - I*a*b^4)*cos(
f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*
b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b
)/b^2))/b^2) - (2*(2*a*b^4 + b^5)*cos(f*x + e)^3 - (5*a^2*b^3 + 7*a*b^4 + 2*b^5)*cos(f*x + e))*sqrt(-b*cos(f*x
+ e)^2 + a + b)*sin(f*x + e))/((a^4*b^4 + 2*a^3*b^5 + a^2*b^6)*f*cos(f*x + e)^4 - 2*(a^5*b^3 + 3*a^4*b^4 + 3*
a^3*b^5 + a^2*b^6)*f*cos(f*x + e)^2 + (a^6*b^2 + 4*a^5*b^3 + 6*a^4*b^4 + 4*a^3*b^5 + a^2*b^6)*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Integral((a + b*sin(e + f*x)**2)**(-5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(-5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(e + f*x)^2)^(5/2),x)

[Out]

int(1/(a + b*sin(e + f*x)^2)^(5/2), x)

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