\(\int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx\) [749]

Optimal result

Integrand size = 27, antiderivative size = 217 \[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {2 d^2 \cos (e+f x)}{(b c-3 d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}-\frac {2 d E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{(b c-3 d) \left (c^2-d^2\right ) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 b \operatorname {EllipticPi}\left (\frac {2 b}{3+b},\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{(3+b) (b c-3 d) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2*d^2*cos(f*x+e)/(-a*d+b*c)/(c^2-d^2)/f/(c+d*sin(f*x+e))^(1/2)+2*d*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/
2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/(-a*d+
b*c)/(c^2-d^2)/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-2*b*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f
*x)*EllipticPi(cos(1/2*e+1/4*Pi+1/2*f*x),2*b/(a+b),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/(a+
b)/(-a*d+b*c)/f/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2881, 3138, 2734, 2732, 12, 2886, 2884} \[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}-\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 b \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{f (a+b) (b c-a d) \sqrt {c+d \sin (e+f x)}} \]

[In]

Int[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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

Rule 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 d^2 \cos (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {2 \int \frac {\frac {1}{2} \left (-a c d+b \left (c^2-d^2\right )\right )-\frac {1}{2} d (b c+a d) \sin (e+f x)-\frac {1}{2} b d^2 \sin ^2(e+f x)}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}} \, dx}{(b c-a d) \left (c^2-d^2\right )} \\ & = -\frac {2 d^2 \cos (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}-\frac {2 \int -\frac {b^2 d \left (c^2-d^2\right )}{2 (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}} \, dx}{b d (b c-a d) \left (c^2-d^2\right )}-\frac {d \int \sqrt {c+d \sin (e+f x)} \, dx}{(b c-a d) \left (c^2-d^2\right )} \\ & = -\frac {2 d^2 \cos (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {b \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}} \, dx}{b c-a d}-\frac {\left (d \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{(b c-a d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \\ & = -\frac {2 d^2 \cos (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}-\frac {2 d E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (b \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{(b c-a d) \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {2 d^2 \cos (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}-\frac {2 d E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 b \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{(a+b) (b c-a d) f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.72 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.78 \[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {\frac {4 d^2 \cos (e+f x)}{\left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}+\frac {-\frac {4 i (b c+3 d) \left ((b c-3 d) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )+3 d \operatorname {EllipticPi}\left (\frac {b (c+d)}{b c-3 d},i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )\right ) \sec (e+f x) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {\frac {d (1+\sin (e+f x))}{-c+d}}}{b (b c-3 d) \sqrt {-\frac {1}{c+d}}}-\frac {2 i \left (-2 b (b c-3 d) (c-d) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right )|\frac {c+d}{c-d}\right )+d \left (-2 (3+b) (b c-3 d) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )+\left (-18+b^2\right ) d \operatorname {EllipticPi}\left (\frac {b (c+d)}{b c-3 d},i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )\right )\right ) \sec (e+f x) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {\frac {d (1+\sin (e+f x))}{-c+d}}}{b (b c-3 d) \sqrt {-\frac {1}{c+d}}}+\frac {2 \left (2 b c^2-6 c d-3 b d^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{3+b},\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{(3+b) \sqrt {c+d \sin (e+f x)}}}{(c-d) (c+d)}}{2 (b c-3 d) f} \]

[In]

Integrate[1/((3 + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

-1/2*((4*d^2*Cos[e + f*x])/((c^2 - d^2)*Sqrt[c + d*Sin[e + f*x]]) + (((-4*I)*(b*c + 3*d)*((b*c - 3*d)*Elliptic
F[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + 3*d*EllipticPi[(b*(c + d))/(b*c
- 3*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)])*Sec[e + f*x]*Sqrt[-((d*(-1
+ Sin[e + f*x]))/(c + d))]*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d)])/(b*(b*c - 3*d)*Sqrt[-(c + d)^(-1)]) - ((2*I)
*(-2*b*(b*c - 3*d)*(c - d)*EllipticE[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)]
+ d*(-2*(3 + b)*(b*c - 3*d)*EllipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d
)] + (-18 + b^2)*d*EllipticPi[(b*(c + d))/(b*c - 3*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]]
, (c + d)/(c - d)]))*Sec[e + f*x]*Sqrt[-((d*(-1 + Sin[e + f*x]))/(c + d))]*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d
)])/(b*(b*c - 3*d)*Sqrt[-(c + d)^(-1)]) + (2*(2*b*c^2 - 6*c*d - 3*b*d^2)*EllipticPi[(2*b)/(3 + b), (-2*e + Pi
- 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/((3 + b)*Sqrt[c + d*Sin[e + f*x]]))/((c - d)*(c
+ d)))/((b*c - 3*d)*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(609\) vs. \(2(276)=552\).

Time = 3.29 (sec) , antiderivative size = 610, normalized size of antiderivative = 2.81

method	result	size
default	\(\frac {\sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\frac {d \left (\frac {2 d \left (\cos ^{2}\left (f x +e \right )\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}+\frac {2 c \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, F\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}+\frac {2 d \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) E\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+F\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{d a -c b}-\frac {2 \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \Pi \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \frac {-\frac {c}{d}+1}{-\frac {c}{d}+\frac {a}{b}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (d a -c b \right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {c}{d}+\frac {a}{b}\right )}\right )}{\cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(610\)

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(d/(a*d-b*c)*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e
)^2)^(1/2)+2*c/(c^2-d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(
f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+
d))^(1/2))+2/(c^2-d^2)*d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(
f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),
((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))-2/(a*d-b*c)*(c/d-1)*((c+d
*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)
*cos(f*x+e)^2)^(1/2)/(-c/d+a/b)*EllipticPi(((c+d*sin(f*x+e))/(c-d))^(1/2),(-c/d+1)/(-c/d+a/b),((c-d)/(c+d))^(1
/2)))/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral(1/((a + b*sin(e + f*x))*(c + d*sin(e + f*x))**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(3+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (a+b\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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