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

Optimal. Leaf size=644 \[ \frac {\sqrt {a+b} (b (c-d)-2 a d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} F\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{d^2 f \sqrt {c+d}}-\frac {\sqrt {a+b} (b c-3 a d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \Pi \left (\frac {(a+b) d}{b (c+d)};\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{d^2 f \sqrt {c+d}}-\frac {b \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{f \sqrt {c+d \sin (e+f x)}}-\frac {b (a-b) \sqrt {a+b} \sqrt {c+d} \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{d f (b c-a d)} \]

[Out]

(b*(c-d)-2*a*d)*EllipticF((c+d)^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),((a+b)*(c-d)/(
a-b)/(c+d))^(1/2))*sec(f*x+e)*(c+d*sin(f*x+e))*(a+b)^(1/2)*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^
(1/2)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a-b)/(c+d*sin(f*x+e)))^(1/2)/d^2/f/(c+d)^(1/2)-(-3*a*d+b*c)*EllipticPi((c+d
)^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),(a+b)*d/b/(c+d),((a+b)*(c-d)/(a-b)/(c+d))^(1
/2))*sec(f*x+e)*(c+d*sin(f*x+e))*(a+b)^(1/2)*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^(1/2)*(-(-a*d+
b*c)*(1+sin(f*x+e))/(a-b)/(c+d*sin(f*x+e)))^(1/2)/d^2/f/(c+d)^(1/2)-(a-b)*b*EllipticE((c+d)^(1/2)*(a+b*sin(f*x
+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),((a+b)*(c-d)/(a-b)/(c+d))^(1/2))*sec(f*x+e)*(c+d*sin(f*x+e))*(a+
b)^(1/2)*(c+d)^(1/2)*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^(1/2)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a-b
)/(c+d*sin(f*x+e)))^(1/2)/d/(-a*d+b*c)/f-b*cos(f*x+e)*(a+b*sin(f*x+e))^(1/2)/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 1.54, antiderivative size = 644, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2821, 3053, 2811, 2998, 2818, 2996} \[ \frac {\sqrt {a+b} (b (c-d)-2 a d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} F\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{d^2 f \sqrt {c+d}}-\frac {\sqrt {a+b} (b c-3 a d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \Pi \left (\frac {(a+b) d}{b (c+d)};\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{d^2 f \sqrt {c+d}}-\frac {b \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{f \sqrt {c+d \sin (e+f x)}}-\frac {b (a-b) \sqrt {a+b} \sqrt {c+d} \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{d f (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]])/(f*Sqrt[c + d*Sin[e + f*x]])) - ((a - b)*b*Sqrt[a + b]*Sqrt[c + d]
*EllipticE[ArcSin[(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]])], ((a + b)*(c
- d))/((a - b)*(c + d))]*Sec[e + f*x]*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d*Sin[e + f*x]))]*Sq
rt[-(((b*c - a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*(c + d*Sin[e + f*x]))/(d*(b*c - a*d)*f)
+ (Sqrt[a + b]*(b*(c - d) - 2*a*d)*EllipticF[ArcSin[(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[
c + d*Sin[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))]*Sec[e + f*x]*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]))
/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*(c +
d*Sin[e + f*x]))/(d^2*Sqrt[c + d]*f) - (Sqrt[a + b]*(b*c - 3*a*d)*EllipticPi[((a + b)*d)/(b*(c + d)), ArcSin[
(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c
+ d))]*Sec[e + f*x]*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*
(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*(c + d*Sin[e + f*x]))/(d^2*Sqrt[c + d]*f)

Rule 2811

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

Rule 2818

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Si
mp[(2*(c + d*Sin[e + f*x])*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c
- a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*EllipticF[ArcSin[Rt[(c + d)/(a + b), 2]*(Sqrt[a +
b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))])/(f*(b*c - a*d)*Rt[(c + d)/(a
+ b), 2]*Cos[e + f*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] && PosQ[(c + d)/(a + b)]

Rule 2821

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[1/(d*(m + n)),
Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a^2*c*d*(m + n) + b*d*(b*c*(m - 1) + a*d*n
) + (a*d*(2*b*c + a*d)*(m + n) - b*d*(a*c - b*d*(m + n - 1)))*Sin[e + f*x] + b*d*(b*c*n + a*d*(2*m + n - 1))*S
in[e + f*x]^2, 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] && LtQ[0, m, 2] && LtQ[-1, n, 2] && NeQ[m + n, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 2996

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*(a + b*Sin[e + f*x])*Sqrt[((b*c - a*d)*(1 + Sin[e + f*
x]))/((c - d)*(a + b*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 - Sin[e + f*x]))/((c + d)*(a + b*Sin[e + f*x])))]*
EllipticE[ArcSin[(Rt[(a + b)/(c + d), 2]*Sqrt[c + d*Sin[e + f*x]])/Sqrt[a + b*Sin[e + f*x]]], ((a - b)*(c + d)
)/((a + b)*(c - d))])/(f*(b*c - a*d)^2*Rt[(a + b)/(c + d), 2]*Cos[e + f*x]), x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(a + b)/(c + d)]

Rule 2998

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 3053

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + (f_.
)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[C/b^2, Int[Sqrt[a + b*Sin[e + f
*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[1/b^2, Int[(A*b^2 - a^2*C + b*(b*B - 2*a*C)*Sin[e + f*x])/((a + b
*Sin[e + f*x])^(3/2)*Sqrt[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*} \int \frac {(a+b \sin (e+f x))^{3/2}}{\sqrt {c+d \sin (e+f x)}} \, dx &=-\frac {b \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{f \sqrt {c+d \sin (e+f x)}}+\frac {\int \frac {\frac {1}{2} d \left (2 a^2 c+b^2 c-a b d\right )+a d (b c+a d) \sin (e+f x)-\frac {1}{2} b d (b c-3 a d) \sin ^2(e+f x)}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx}{d}\\ &=-\frac {b \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{f \sqrt {c+d \sin (e+f x)}}+\frac {\int \frac {\frac {1}{2} b c^2 d (b c-3 a d)+\frac {1}{2} d^3 \left (2 a^2 c+b^2 c-a b d\right )+d \left (b c d (b c-3 a d)+a d^2 (b c+a d)\right ) \sin (e+f x)}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx}{d^3}-\frac {(b (b c-3 a d)) \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a+b \sin (e+f x)}} \, dx}{2 d^2}\\ &=-\frac {b \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{f \sqrt {c+d \sin (e+f x)}}-\frac {\sqrt {a+b} (b c-3 a d) \Pi \left (\frac {(a+b) d}{b (c+d)};\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{d^2 \sqrt {c+d} f}+\frac {(b (c+d) (b c-a d)) \int \frac {1+\sin (e+f x)}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx}{2 d}+\frac {\left (\frac {1}{2} b c^2 d (b c-3 a d)+\frac {1}{2} d^3 \left (2 a^2 c+b^2 c-a b d\right )-d \left (b c d (b c-3 a d)+a d^2 (b c+a d)\right )\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{(c-d) d^3}\\ &=-\frac {b \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{f \sqrt {c+d \sin (e+f x)}}-\frac {(a-b) b \sqrt {a+b} \sqrt {c+d} E\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{d (b c-a d) f}+\frac {\sqrt {a+b} (b (c-d)-2 a d) F\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{d^2 \sqrt {c+d} f}-\frac {\sqrt {a+b} (b c-3 a d) \Pi \left (\frac {(a+b) d}{b (c+d)};\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{d^2 \sqrt {c+d} f}\\ \end {align*}

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Mathematica [C]  time = 32.69, size = 222963, normalized size = 346.22 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 2.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {d \sin \left (f x + e\right ) + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 10.60, size = 529691, normalized size = 822.50 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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