3.745 \(\int \frac{(c+d \sin (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx\)
Optimal. Leaf size=296 \[ -\frac{2 d \left (-3 a^2 d^2+6 a b c d+b^2 \left (-\left (2 c^2+d^2\right )\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 b^3 f \sqrt{c+d \sin (e+f x)}}+\frac{2 d (7 b c-3 a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 b^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 (b c-a d)^3 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^3 f (a+b) \sqrt{c+d \sin (e+f x)}}-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f} \]
[Out]
(-2*d^2*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*b*f) + (2*d*(7*b*c - 3*a*d)*EllipticE[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*b^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*d*(6*a*b*c*d - 3*a^2*
d^2 - b^2*(2*c^2 + d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*b
^3*f*Sqrt[c + d*Sin[e + f*x]]) + (2*(b*c - a*d)^3*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^3*(a + b)*f*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 1.08829, antiderivative size = 296, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{2793, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac{2 d \left (-3 a^2 d^2+6 a b c d+b^2 \left (-\left (2 c^2+d^2\right )\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 b^3 f \sqrt{c+d \sin (e+f x)}}+\frac{2 d (7 b c-3 a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 b^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 (b c-a d)^3 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^3 f (a+b) \sqrt{c+d \sin (e+f x)}}-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sin[e + f*x])^(5/2)/(a + b*Sin[e + f*x]),x]
[Out]
(-2*d^2*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*b*f) + (2*d*(7*b*c - 3*a*d)*EllipticE[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*b^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*d*(6*a*b*c*d - 3*a^2*
d^2 - b^2*(2*c^2 + d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*b
^3*f*Sqrt[c + d*Sin[e + f*x]]) + (2*(b*c - a*d)^3*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^3*(a + b)*f*Sqrt[c + d*Sin[e + f*x]])
Rule 2793
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*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] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))
Rule 3059
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]
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 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 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 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 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 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]
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]
Rubi steps
\begin{align*} \int \frac{(c+d \sin (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx &=-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f}+\frac{2 \int \frac{\frac{1}{2} \left (3 b c^3+a d^3\right )-\frac{1}{2} d \left (2 a c d-b \left (9 c^2+d^2\right )\right ) \sin (e+f x)+\frac{1}{2} d^2 (7 b c-3 a d) \sin ^2(e+f x)}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{3 b}\\ &=-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f}-\frac{2 \int \frac{\frac{1}{2} d \left (a c d (7 b c-3 a d)-b \left (3 b c^3+a d^3\right )\right )+\frac{1}{2} d^2 \left (6 a b c d-3 a^2 d^2-b^2 \left (2 c^2+d^2\right )\right ) \sin (e+f x)}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{3 b^2 d}+\frac{(d (7 b c-3 a d)) \int \sqrt{c+d \sin (e+f x)} \, dx}{3 b^2}\\ &=-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f}+\frac{(b c-a d)^3 \int \frac{1}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{b^3}-\frac{\left (d \left (6 a b c d-3 a^2 d^2-b^2 \left (2 c^2+d^2\right )\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 b^3}+\frac{\left (d (7 b c-3 a d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{3 b^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}\\ &=-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f}+\frac{2 d (7 b c-3 a 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)}}{3 b^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left ((b c-a d)^3 \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^3 \sqrt{c+d \sin (e+f x)}}-\frac{\left (d \left (6 a b c d-3 a^2 d^2-b^2 \left (2 c^2+d^2\right )\right ) \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}{3 b^3 \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 b f}+\frac{2 d (7 b c-3 a 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)}}{3 b^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 d \left (6 a b c d-3 a^2 d^2-b^2 \left (2 c^2+d^2\right )\right ) 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}}}{3 b^3 f \sqrt{c+d \sin (e+f x)}}+\frac{2 (b c-a d)^3 \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^3 (a+b) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 5.88064, size = 606, normalized size = 2.05 \[ \frac{\frac{2 i (3 a d-7 b c) \sec (e+f x) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{\frac{d (\sin (e+f x)+1)}{d-c}} \left (d \left (d \left (b^2-2 a^2\right ) \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+2 (a+b) (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )-2 b (c-d) (b c-a d) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )}{b^2 \sqrt{-\frac{1}{c+d}} (b c-a d)}-\frac{2 \left (-a d^3+6 b c^3+7 b c d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )}{(a+b) \sqrt{c+d \sin (e+f x)}}+\frac{4 i \left (b \left (9 c^2+d^2\right )-2 a c d\right ) \sec (e+f x) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{\frac{d (\sin (e+f x)+1)}{d-c}} \left ((a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )-a d \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )}{b \sqrt{-\frac{1}{c+d}} (b c-a d)}-4 d^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{6 b f} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*Sin[e + f*x])^(5/2)/(a + b*Sin[e + f*x]),x]
[Out]
(((4*I)*(-2*a*c*d + b*(9*c^2 + d^2))*((-(b*c) + a*d)*EllipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e
+ f*x]]], (c + d)/(c - d)] - 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*Sqrt[-(c + d)^(-1)]*(b*c - a*d)) + ((2*I)*(-7*b*c + 3*a*d)*(-2*b*(c - d)*(b*c - a*d)*Ellip
ticE[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + d*(2*(a + b)*(-(b*c) + a*d)*E
llipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + (-2*a^2 + b^2)*d*Elliptic
Pi[(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)
]*(b*c - a*d)) - 4*d^2*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]] - (2*(6*b*c^3 + 7*b*c*d^2 - a*d^3)*EllipticPi[(2*
b)/(a + b), (-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/((a + b)*Sqrt[c + d*Sin[
e + f*x]]))/(6*b*f)
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Maple [B] time = 3.692, size = 1190, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x)
[Out]
(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(d/b^3*(b^2*d^2*(-2/3/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2/3*(c
/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-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))-4/3*c/d*(c/d-1)*((
c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-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*b*d^2+3*b^2*c*d)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*
(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-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^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)*((-sin(f*x+e
)-1)*d/(c-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))-6*a*b*c*d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d
/(c-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))+6*c^2*b^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-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*(-a^3*d^3+3*a^2*b*c*d^2-3*a*b^2*c^2*d+b^3*c^3)/b^4*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))
/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-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
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")
[Out]
integrate((d*sin(f*x + e) + c)^(5/2)/(b*sin(f*x + e) + a), x)
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Fricas [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((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")
[Out]
Timed out
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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((c+d*sin(f*x+e))**(5/2)/(a+b*sin(f*x+e)),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="giac")
[Out]
Exception raised: TypeError