3.584 \(\int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=183 \[ \frac{2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}} \]
[Out]
(-2*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(5/
2)*f) + (2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) + (2*
a^3*(c - d)*(3*c + 7*d)*Cos[e + f*x])/(3*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 0.443984, antiderivative size = 183, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used =
{2762, 2980, 2775, 205} \[ \frac{2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(5/2),x]
[Out]
(-2*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(5/
2)*f) + (2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) + (2*
a^3*(c - d)*(3*c + 7*d)*Cos[e + f*x])/(3*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
Rule 2762
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Rule 2980
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Rule 2775
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{(2 a) \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (c-7 d)-\frac{3}{2} a (c+d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac{2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{a^2 \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{d^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac{2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{d^2 f}\\ &=-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac{2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 7.88622, size = 261, normalized size = 1.43 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (\frac{2 (c-d) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 c^2+4 d (c+2 d) \sin (e+f x)+8 c d+d^2\right )}{3 d^2 (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )-\log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )}{d^{5/2}}\right )}{f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(5/2),x]
[Out]
((a*(1 + Sin[e + f*x]))^(5/2)*((2*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]]
+ ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - Log[Sqrt[2]*Sqrt[d]*Cos[(2*
e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]])/d^(5/2) + (2*(c - d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*
c^2 + 8*c*d + d^2 + 4*d*(c + 2*d)*Sin[e + f*x]))/(3*d^2*(c + d)^2*(c + d*Sin[e + f*x])^(3/2))))/(f*(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2])^5)
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Maple [B] time = 0.435, size = 16223, normalized size = 88.7 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x)
[Out]
result too large to display
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(5/2), x)
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Fricas [B] time = 5.4359, size = 5053, normalized size = 27.61 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[-1/12*(3*(a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^
4)*cos(f*x + e)^3 - (2*a^2*c^3*d + 5*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + (a^2*c^4 + 2*a^2*c^
3*d + 2*a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e) + (a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d
^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 2*a^2*c^2*d^2 + a^2*c*d^3
)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a
*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a*c^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e
)^3 - 32*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4*cos(f*x + e)^4 - c^3*d + 17*
c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*(5*c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)
^2 - (c^3*d - 7*c^2*d^2 + 31*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 17*c^2*d^2 + 59*c
*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x +
e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 476*
a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 +
a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*cos(f*x + e)^2 + 32*(a*c
^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) + 8*
(3*a^2*c^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^3 + 4*(a^2*c^2*d + a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e)^2 + (3*
a^2*c^3 + 5*a^2*c^2*d - 7*a^2*c*d^2 - a^2*d^3)*cos(f*x + e) - (3*a^2*c^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^
3 - 4*(a^2*c^2*d + a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x
+ e) + c))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e
)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*
c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e)
- (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e)), -1/6*(3*(a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c
^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^3 - (2*a^2*c^3*d + 5*a^2*c
^2*d^2 + 4*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + (a^2*c^4 + 2*a^2*c^3*d + 2*a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^
4)*cos(f*x + e) + (a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3
+ a^2*d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 2*a^2*c^2*d^2 + a^2*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a/d)*ar
ctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2 + 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq
rt(d*sin(f*x + e) + c)*sqrt(a/d)/(2*a*d^2*cos(f*x + e)^3 - (3*a*c*d - a*d^2)*cos(f*x + e)*sin(f*x + e) - (a*c^
2 - a*c*d + 2*a*d^2)*cos(f*x + e))) + 4*(3*a^2*c^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^3 + 4*(a^2*c^2*d + a^2
*c*d^2 - 2*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 5*a^2*c^2*d - 7*a^2*c*d^2 - a^2*d^3)*cos(f*x + e) - (3*a^2*c
^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^3 - 4*(a^2*c^2*d + a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 +
5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)
- (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*
d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e))]
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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((a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(5/2), x)