3.4.61 \(\int \frac {1}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [361]

Optimal. Leaf size=101 \[ \frac {b \cos (e+f x) \sin (e+f x)}{a (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{a (a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \]

[Out]

b*cos(f*x+e)*sin(f*x+e)/a/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)+(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/(a+b)/f/(1+b*sin(f*x+e)^2/a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3263, 21, 3257, 3256} \begin {gather*} \frac {b \sin (e+f x) \cos (e+f x)}{a f (a+b) \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{a f (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^(-3/2),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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

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Mathematica [A]
time = 0.11, size = 90, normalized size = 0.89 \begin {gather*} \frac {2 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} b \sin (2 (e+f x))}{2 a (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x]^2)^(-3/2),x]

[Out]

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

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Maple [A]
time = 8.12, size = 103, normalized size = 1.02

method	result	size
default	\(\frac {a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) b}{a \left (a +b \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

(a*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))+sin(f*x+e)*cos(
f*x+e)^2*b)/a/(a+b)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(-3/2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.17, size = 938, normalized size = 9.29 \begin {gather*} -\frac {2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, {\left (i \, b^{3} \cos \left (f x + e\right )^{2} - i \, a b^{2} - i \, b^{3}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left (2 i \, a^{2} b + 3 i \, a b^{2} + i \, b^{3} + {\left (-2 i \, a b^{2} - i \, b^{3}\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 (-i \, b^{3} \cos \left (f x + e\right )^{2} + i \, a b^{2} + i \, b^{3}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left (-2 i \, a^{2} b - 3 i \, a b^{2} - i \, b^{3} + {\left (2 i \, a b^{2} + i \, b^{3}\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}}) + 2 \, {\left (2 \, {\left (-i \, a^{2} b - 2 i \, a b^{2} - i \, b^{3} + {\left (i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (2 i \, a^{3} + 3 i \, a^{2} b + i \, a b^{2} + {\left (-2 i \, a^{2} b - i \, a b^{2}\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}}) + 2 \, {\left (2 \, {\left (i \, a^{2} b + 2 i \, a b^{2} + i \, b^{3} + {\left (-i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (-2 i \, a^{3} - 3 i \, a^{2} b - i \, a b^{2} + {\left (2 i \, a^{2} b + i \, a b^{2}\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}})}{2 \, {\left ({\left (a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

-1/2*(2*sqrt(-b*cos(f*x + e)^2 + a + b)*b^3*cos(f*x + e)*sin(f*x + e) - (2*(I*b^3*cos(f*x + e)^2 - I*a*b^2 - I
*b^3)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (2*I*a^2*b + 3*I*a*b^2 + I*b^3 + (-2*I*a*b^2 - I*b^3)*cos(f*x + e)^2)*s
qrt(-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*(-I*b^3*cos(f*x + e)^2 + I*a*b^2 + I*b^3)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (-2*I*a^2*b - 3*I*a*b^2 - I*b^
3 + (2*I*a*b^2 + I*b^3)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arc
sin(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^2*b - 2*I*a*b^2 - I*b^3 + (I*a*b^2 + I*b^3)*cos(f*x + e)
^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) + (2*I*a^3 + 3*I*a^2*b + I*a*b^2 + (-2*I*a^2*b - I*a*b^2)*cos(f*x + e)^2)*s
qrt(-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*(I*a^2*b + 2*I*a*b^2 + I*b^3 + (-I*a*b^2 - I*b^3)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) + (-2*
I*a^3 - 3*I*a^2*b - I*a*b^2 + (2*I*a^2*b + I*a*b^2)*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))/((a^2*b^3 + a*b^4)*f*cos(f*x + e)^2 - (a
^3*b^2 + 2*a^2*b^3 + a*b^4)*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 {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral((a + b*sin(e + f*x)**2)**(-3/2), 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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(1/(a + b*sin(e + f*x)^2)^(3/2), x)

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