3.575 \(\int \frac{(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=117 \[ \frac{2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{d^{3/2} f} \]
[Out]
(-2*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(3/
2)*f) + (2*a^2*(c - d)*Cos[e + f*x])/(d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 0.213653, antiderivative size = 117, 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, 21, 2775, 205} \[ \frac{2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{d^{3/2} f} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(3/2),x]
[Out]
(-2*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(3/
2)*f) + (2*a^2*(c - d)*Cos[e + f*x])/(d*(c + d)*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 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
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))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{(2 a) \int \frac{-\frac{1}{2} a (c+d)-\frac{1}{2} a (c+d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{a \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{d}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{\left (2 a^2\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 f}\\ &=-\frac{2 a^{3/2} \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^{3/2} f}+\frac{2 a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [B] time = 7.27412, size = 377, normalized size = 3.22 \[ \frac{(a (\sin (e+f x)+1))^{3/2} \left (-2 c \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )+2 c \sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )+2 (c+d) \sqrt{c+d \sin (e+f x)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )-d \sqrt{c+d \sin (e+f x)} \log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-c \sqrt{c+d \sin (e+f x)} \log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+(c+d) \sqrt{c+d \sin (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )+2 d^{3/2} \sin \left (\frac{1}{2} (e+f x)\right )-2 d^{3/2} \cos \left (\frac{1}{2} (e+f x)\right )\right )}{d^{3/2} f (c+d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(3/2),x]
[Out]
((a*(1 + Sin[e + f*x]))^(3/2)*(2*c*Sqrt[d]*Cos[(e + f*x)/2] - 2*d^(3/2)*Cos[(e + f*x)/2] - 2*c*Sqrt[d]*Sin[(e
+ f*x)/2] + 2*d^(3/2)*Sin[(e + f*x)/2] + 2*(c + d)*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c +
d*Sin[e + f*x]]]*Sqrt[c + d*Sin[e + f*x]] + (c + d)*ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[
c + d*Sin[e + f*x]]]*Sqrt[c + d*Sin[e + f*x]] - c*Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*S
in[e + f*x]]]*Sqrt[c + d*Sin[e + f*x]] - d*Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e +
f*x]]]*Sqrt[c + d*Sin[e + f*x]]))/(d^(3/2)*(c + d)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c + d*Sin[e
+ f*x]])
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Maple [B] time = 0.428, size = 6627, normalized size = 56.6 \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))^(3/2)/(c+d*sin(f*x+e))^(3/2),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{3}{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))^(3/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^(3/2)/(d*sin(f*x + e) + c)^(3/2), x)
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Fricas [B] time = 7.01481, size = 3043, normalized size = 26.01 \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))^(3/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
[-1/4*((a*c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a
*c*d + a*d^2 + (a*c*d + a*d^2)*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*(a*c - a*d + (a*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f*x + e))*sqrt(a*sin(f*x
+ e) + a)*sqrt(d*sin(f*x + e) + c))/((c*d^2 + d^3)*f*cos(f*x + e)^2 - (c^2*d + c*d^2)*f*cos(f*x + e) - (c^2*d
+ 2*c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*cos(f*x + e) + (c^2*d + 2*c*d^2 + d^3)*f)*sin(f*x + e)), -1/2*((a*c^2 +
2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d^2 + (
a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a/d)*arctan(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)*sqrt(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*(a*c - a*d + (a
*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((c*d^2
+ d^3)*f*cos(f*x + e)^2 - (c^2*d + c*d^2)*f*cos(f*x + e) - (c^2*d + 2*c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*cos(f*
x + e) + (c^2*d + 2*c*d^2 + d^3)*f)*sin(f*x + e))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}{\left (c + d \sin{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**(3/2),x)
[Out]
Integral((a*(sin(e + f*x) + 1))**(3/2)/(c + d*sin(e + f*x))**(3/2), x)
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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{3}{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))^(3/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^(3/2)/(d*sin(f*x + e) + c)^(3/2), x)