3.507 \(\int (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=154 \[ -\frac{b \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{a (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 (2 a+b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.167102, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3180, 3172, 3178, 3177, 3183, 3182} \[ -\frac{b \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{a (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 (2 a+b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

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

Rule 3180

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

Rule 3172

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 3178

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 3177

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

Rule 3183

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 3182

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

Rubi steps

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

Mathematica [A]  time = 0.767113, size = 156, normalized size = 1.01 \[ \frac{b \sin (2 (e+f x)) (-2 a+b \cos (2 (e+f x))-b)-2 \sqrt{2} a (a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )+4 \sqrt{2} a (2 a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{6 \sqrt{2} f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a*(2*a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] - 2*Sqrt[2]*a*(a + b)
*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + b*(-2*a - b + b*Cos[2*(e + f*x)])*Sin[2*(
e + f*x)])/(6*Sqrt[2]*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [A]  time = 1.317, size = 266, normalized size = 1.7 \begin{align*}{\frac{1}{f\cos \left ( fx+e \right ) } \left ( -{\frac{{a}^{2}}{3}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }-{\frac{ab}{3}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }+{\frac{4\,{a}^{2}}{3}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }+{\frac{2\,ab}{3}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }+{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{5}{b}^{2}}{3}}+{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}ab}{3}}-{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}{b}^{2}}{3}}-{\frac{\sin \left ( fx+e \right ) ab}{3}} \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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^2-1/3*a*(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+4/3*(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+2/3*(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*b+1/3*sin(f*x+e)^5*b^2+1/3*sin(f*x+e)^3*a*b-1/3*sin(f*x+e)^3
*b^2-1/3*sin(f*x+e)*a*b)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral((-b*cos(f*x + e)^2 + a + b)^(3/2), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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