\(\int \sin ^2(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [139]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 218 \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {(3 a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}+\frac {\left (3 a^2+13 a b+8 b^2\right ) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{15 b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a (a+b) (3 a+4 b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-1/5*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/f-1/15*(3*a+4*b)*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^
(1/2)/f+1/15*(3*a^2+13*a*b+8*b^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)/b/f/(1+b*sin(f*x+e)^2/a)^(1/2)-1/15*a*(a+b)*(3*a+4*b)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*Elliptic
F(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/b/f/(a+b*sin(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3249, 3251, 3257, 3256, 3262, 3261} \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (3 a^2+13 a b+8 b^2\right ) \sqrt {a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{15 b f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\sin (e+f x) \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}-\frac {(3 a+4 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 f}-\frac {a (a+b) (3 a+4 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )}{15 b f \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

-1/15*((3*a + 4*b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/f - (Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x]^2)^(3/2))/(5*f) + ((3*a^2 + 13*a*b + 8*b^2)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2
])/(15*b*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (a*(a + b)*(3*a + 4*b)*EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b*Sin
[e + f*x]^2)/a])/(15*b*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3249

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

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 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}+\frac {1}{5} \int \sqrt {a+b \sin ^2(e+f x)} \left (a+(3 a+4 b) \sin ^2(e+f x)\right ) \, dx \\ & = -\frac {(3 a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}+\frac {1}{15} \int \frac {2 a (3 a+2 b)+\left (3 a^2+13 a b+8 b^2\right ) \sin ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx \\ & = -\frac {(3 a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}-\frac {(a (a+b) (3 a+4 b)) \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx}{15 b}+\frac {\left (3 a^2+13 a b+8 b^2\right ) \int \sqrt {a+b \sin ^2(e+f x)} \, dx}{15 b} \\ & = -\frac {(3 a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}+\frac {\left (\left (3 a^2+13 a b+8 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}\right ) \int \sqrt {1+\frac {b \sin ^2(e+f x)}{a}} \, dx}{15 b \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {\left (a (a+b) (3 a+4 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}{15 b \sqrt {a+b \sin ^2(e+f x)}} \\ & = -\frac {(3 a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 f}+\frac {\left (3 a^2+13 a b+8 b^2\right ) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{15 b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a (a+b) (3 a+4 b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.92 \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {16 a \left (3 a^2+13 a b+8 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-16 a \left (3 a^2+7 a b+4 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\sqrt {2} b \left (48 a^2+68 a b+25 b^2-4 b (9 a+7 b) \cos (2 (e+f x))+3 b^2 \cos (4 (e+f x))\right ) \sin (2 (e+f x))}{240 b f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

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

[Out]

(16*a*(3*a^2 + 13*a*b + 8*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] - 16*a*(3*a^2
 + 7*a*b + 4*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(48*a^2 + 68*a
*b + 25*b^2 - 4*b*(9*a + 7*b)*Cos[2*(e + f*x)] + 3*b^2*Cos[4*(e + f*x)])*Sin[2*(e + f*x)])/(240*b*f*Sqrt[2*a +
 b - b*Cos[2*(e + f*x)]])

Maple [A] (verified)

Time = 3.54 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.97

method result size
default \(-\frac {-3 b^{3} \left (\sin ^{7}\left (f x +e \right )\right )-9 a \,b^{2} \left (\sin ^{5}\left (f x +e \right )\right )-b^{3} \left (\sin ^{5}\left (f x +e \right )\right )+3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+7 a^{2} \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b +4 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}-3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-13 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b -8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}-6 a^{2} b \left (\sin ^{3}\left (f x +e \right )\right )+5 a \,b^{2} \left (\sin ^{3}\left (f x +e \right )\right )+4 b^{3} \left (\sin ^{3}\left (f x +e \right )\right )+6 a^{2} b \sin \left (f x +e \right )+4 a \,b^{2} \sin \left (f x +e \right )}{15 b \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) \(429\)

[In]

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

[Out]

-1/15*(-3*b^3*sin(f*x+e)^7-9*a*b^2*sin(f*x+e)^5-b^3*sin(f*x+e)^5+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^3+7*a^2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*Ellipt
icF(sin(f*x+e),(-1/a*b)^(1/2))*b+4*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^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^3
-13*(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-8*(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^2-6*a^2*b*sin(f*x+e)^3+5*a*b^2
*sin(f*x+e)^3+4*b^3*sin(f*x+e)^3+6*a^2*b*sin(f*x+e)+4*a*b^2*sin(f*x+e))/b/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/
f

Fricas [F]

\[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{2} \,d x } \]

[In]

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

[Out]

integral((b*cos(f*x + e)^4 - (a + 2*b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b), x)

Sympy [F(-1)]

Timed out. \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int {\sin \left (e+f\,x\right )}^2\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

[In]

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

[Out]

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