3.583 \(\int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=180 \[ -\frac{a^3 (3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{d^2 f (c+d) \sqrt{a \sin (e+f x)+a}}+\frac{a^{5/2} (3 c-5 d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{d f (c+d) \sqrt{c+d \sin (e+f x)}} \]

[Out]

(a^(5/2)*(3*c - 5*d)*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]])/(d*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])
- (a^3*(3*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(d^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.438198, antiderivative size = 180, 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, 2981, 2775, 205} \[ -\frac{a^3 (3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{d^2 f (c+d) \sqrt{a \sin (e+f x)+a}}+\frac{a^{5/2} (3 c-5 d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{d f (c+d) \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(3/2),x]

[Out]

(a^(5/2)*(3*c - 5*d)*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]])/(d*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])
- (a^3*(3*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(d^2*(c + d)*f*Sqrt[a + a*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 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[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))^{3/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{(2 a) \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (c-3 d)-\frac{1}{2} a (3 c-d) \sin (e+f x)\right )}{\sqrt{c+d \sin (e+f x)}} \, dx}{d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{a^3 (3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{d^2 (c+d) f \sqrt{a+a \sin (e+f x)}}-\frac{\left (a^2 (3 c-5 d)\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{2 d^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{a^3 (3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{d^2 (c+d) f \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^3 (3 c-5 d)\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{a^{5/2} (3 c-5 d) \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)}}{d (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{a^3 (3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{d^2 (c+d) f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.00875, size = 263, normalized size = 1.46 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (\frac{(5 d-3 c) \left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )-\log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )\right )}{d^{5/2}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 c^2+d (c+d) \sin (e+f x)-3 c d+2 d^2\right )}{d^2 (c+d) \sqrt{c+d \sin (e+f x)}}\right )}{2 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(3/2),x]

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(((-3*c + 5*d)*(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]*S
qrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(5/2) - (2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])*(3*c^2 - 3*c*d + 2*d^2 + d*(c + d)*Sin[e + f*x]))/(d^2*(c + d)*Sqrt[c + d*Sin[e + f*x]])))/(2*f*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [F]  time = 0.252, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \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))^(3/2),x)

[Out]

int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(3/2),x)

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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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(3/2), x)

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Fricas [B]  time = 5.57539, size = 3758, normalized size = 20.88 \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))^(3/2),x, algorithm="fricas")

[Out]

[1/8*((3*a^2*c^3 + a^2*c^2*d - 7*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2*c^2*d - 2*a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e)^
2 + (3*a^2*c^3 - 2*a^2*c^2*d - 5*a^2*c*d^2)*cos(f*x + e) + (3*a^2*c^3 + a^2*c^2*d - 7*a^2*c*d^2 - 5*a^2*d^3 +
(3*a^2*c^2*d - 2*a^2*c*d^2 - 5*a^2*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^2 - 4*a^2*c*d + a^2*d^2 + (a^2*c*d + a^2*d^2)*cos(f*x + e)^2 +
(3*a^2*c^2 - 3*a^2*c*d + 2*a^2*d^2)*cos(f*x + e) - (3*a^2*c^2 - 4*a^2*c*d + a^2*d^2 - (a^2*c*d + a^2*d^2)*cos(
f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((c*d^3 + d^4)*f*cos(f*x + e)^2 - (
c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^4)*f - ((c*d^3 + d^4)*f*cos(f*x + e) + (c^2*d^2 + 2*c
*d^3 + d^4)*f)*sin(f*x + e)), 1/4*((3*a^2*c^3 + a^2*c^2*d - 7*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2*c^2*d - 2*a^2*c*d
^2 - 5*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 - 2*a^2*c^2*d - 5*a^2*c*d^2)*cos(f*x + e) + (3*a^2*c^3 + a^2*c^2*d
- 7*a^2*c*d^2 - 5*a^2*d^3 + (3*a^2*c^2*d - 2*a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a/d)*arc
tan(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)*sqr
t(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^2 - 4*a^2*c*d + a^2*d^2 + (a^2*c*d + a^2*d^2)*cos(f*x + e)^2 +
(3*a^2*c^2 - 3*a^2*c*d + 2*a^2*d^2)*cos(f*x + e) - (3*a^2*c^2 - 4*a^2*c*d + a^2*d^2 - (a^2*c*d + a^2*d^2)*cos
(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((c*d^3 + d^4)*f*cos(f*x + e)^2 -
(c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^4)*f - ((c*d^3 + d^4)*f*cos(f*x + e) + (c^2*d^2 + 2*
c*d^3 + d^4)*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))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(3/2), x)