3.511 \(\int \frac{(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=237 \[ -\frac{(c+3 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 a^2 f (\sin (e+f x)+1)}+\frac{(c+d) (c+2 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a \sin (e+f x)+a)^2} \]
[Out]
-((c + 3*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*a^2*f*(1 + Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*Sqrt[c
+ d*Sin[e + f*x]])/(3*f*(a + a*Sin[e + f*x])^2) - ((c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqr
t[c + d*Sin[e + f*x]])/(3*a^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c + d)*(c + 2*d)*EllipticF[(e - Pi/2 +
f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a^2*f*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 0.537778, antiderivative size = 237, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used =
{2765, 2978, 2752, 2663, 2661, 2655, 2653} \[ -\frac{(c+3 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 a^2 f (\sin (e+f x)+1)}+\frac{(c+d) (c+2 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a \sin (e+f x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^2,x]
[Out]
-((c + 3*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*a^2*f*(1 + Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*Sqrt[c
+ d*Sin[e + f*x]])/(3*f*(a + a*Sin[e + f*x])^2) - ((c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqr
t[c + d*Sin[e + f*x]])/(3*a^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c + d)*(c + 2*d)*EllipticF[(e - Pi/2 +
f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a^2*f*Sqrt[c + d*Sin[e + f*x]])
Rule 2765
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Rule 2978
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])
Rule 2752
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*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]
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 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]
Rubi steps
\begin{align*} \int \frac{(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx &=-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac{\int \frac{-\frac{1}{2} a \left (2 c^2+5 c d-d^2\right )-\frac{1}{2} a d (c+5 d) \sin (e+f x)}{(a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{3 a^2}\\ &=-\frac{(c+3 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}+\frac{\int \frac{a^2 (c-d) d^2-\frac{1}{2} a^2 (c-d) d (c+3 d) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 a^4 (c-d)}\\ &=-\frac{(c+3 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}+\frac{((c+d) (c+2 d)) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{6 a^2}-\frac{(c+3 d) \int \sqrt{c+d \sin (e+f x)} \, dx}{6 a^2}\\ &=-\frac{(c+3 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac{\left ((c+3 d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{6 a^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left ((c+d) (c+2 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}{6 a^2 \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{(c+3 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac{(c+3 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 a^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(c+d) (c+2 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}}}{3 a^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 2.7908, size = 283, normalized size = 1.19 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (-2 d^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-(c+3 d) (c+d \sin (e+f x))+\frac{(c+d \sin (e+f x)) \left ((3 c+5 d) \sin \left (\frac{1}{2} (e+f x)\right )-(c+3 d) \cos \left (\frac{3}{2} (e+f x)\right )+4 d \cos \left (\frac{1}{2} (e+f x)\right )\right )}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}+(c+3 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )\right )}{3 a^2 f (\sin (e+f x)+1)^2 \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^2,x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-((c + 3*d)*(c + d*Sin[e + f*x])) + ((4*d*Cos[(e + f*x)/2] - (c + 3*
d)*Cos[(3*(e + f*x))/2] + (3*c + 5*d)*Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2])^3 - 2*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + (c + 3*d)
*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])
*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(3*a^2*f*(1 + Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] time = 4.859, size = 1049, normalized size = 4.4 \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))^(3/2)/(a+a*sin(f*x+e))^2,x)
[Out]
(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^2*(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))+(c^2-2*c*d+d^2)*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+sin(
f*x+e))^2-1/3*(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^2*(c-3*d)/((-d*sin(f*x+e)-c)*(-1+sin(f*x+e))
*(1+sin(f*x+e)))^(1/2)+2*d^2/(3*c^2-6*c*d+3*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))-1/3*d*(c-3*d)/(c-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)*((-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*d*(c-d)*(-(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)/((-d*sin(f*x+e)-c)*(-1+sin(f*x+e))*(1+sin(f
*x+e)))^(1/2)-2*d/(2*c-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)*((-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))-d/(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)))))/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{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a)^2, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
integral(-(d*sin(f*x + e) + c)^(3/2)/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)
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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))**(3/2)/(a+a*sin(f*x+e))**2,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")
[Out]
integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a)^2, x)