3.746 \(\int \frac {(c+d \sin (e+f x))^{3/2}}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=229 \[ \frac {2 d (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{b^2 f \sqrt {c+d \sin (e+f x)}}+\frac {2 (b c-a d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \Pi \left (\frac {2 b}{a+b};\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{b^2 f (a+b) \sqrt {c+d \sin (e+f x)}}+\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{b f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2804, 2655, 2653, 2803, 2663, 2661, 2807, 2805} \[ \frac {2 d (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{b^2 f \sqrt {c+d \sin (e+f x)}}+\frac {2 (b c-a d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \Pi \left (\frac {2 b}{a+b};\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{b^2 f (a+b) \sqrt {c+d \sin (e+f x)}}+\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{b f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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

Rule 2803

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

Rule 2804

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[b
/d, Int[Sqrt[a + b*Sin[e + f*x]], x], x] - Dist[(b*c - a*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]/(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] && NeQ[c^2 - d^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 2807

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*
Sin[e + f*x])/(c + d)]), 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]

Rubi steps

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

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Mathematica [C]  time = 3.96, size = 242, normalized size = 1.06 \[ \frac {2 i \sec (e+f x) \sqrt {-\frac {d (\sin (e+f x)-1)}{c+d}} \sqrt {-\frac {d (\sin (e+f x)+1)}{c-d}} \left ((a d+b (d-2 c)) F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right )|\frac {c+d}{c-d}\right )+(b c-a d) \Pi \left (\frac {b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right )|\frac {c+d}{c-d}\right )+b (c-d) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right )|\frac {c+d}{c-d}\right )\right )}{b^2 f \sqrt {-\frac {1}{c+d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*(b*(c - d)*EllipticE[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + (a*d +
b*(-2*c + d))*EllipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + (b*c - a*
d)*EllipticPi[(b*(c + d))/(b*c - a*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^2*Sqrt[-
(c + d)^(-1)]*f)

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fricas [F]  time = 24.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{b \sin \left (f x + e\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{b \sin \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.52, size = 391, normalized size = 1.71 \[ -\frac {2 \left (\EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) b c +\EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) b d +a \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) d -2 c b \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )-\EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) b d -\EllipticPi \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, -\frac {\left (c -d \right ) b}{d a -c b}, \sqrt {\frac {c -d}{c +d}}\right ) a d +\EllipticPi \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, -\frac {\left (c -d \right ) b}{d a -c b}, \sqrt {\frac {c -d}{c +d}}\right ) b c \right ) \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \sqrt {-\frac {\left (\sin \left (f x +e \right )-1\right ) d}{c +d}}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \left (c -d \right )}{b^{2} \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{b \sin \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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