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

Optimal. Leaf size=98 \[ \frac{2 a^{3/2} (c-d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt{c+d}}-\frac{2 a^2 \cos (e+f x)}{d f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.1998, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2763, 21, 2773, 208} \[ \frac{2 a^{3/2} (c-d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt{c+d}}-\frac{2 a^2 \cos (e+f x)}{d f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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 - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx &=-\frac{2 a^2 \cos (e+f x)}{d f \sqrt{a+a \sin (e+f x)}}+\frac{2 \int \frac{-\frac{1}{2} a^2 (c-d)-\frac{1}{2} a^2 (c-d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{d}\\ &=-\frac{2 a^2 \cos (e+f x)}{d f \sqrt{a+a \sin (e+f x)}}-\frac{(a (c-d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{d}\\ &=-\frac{2 a^2 \cos (e+f x)}{d f \sqrt{a+a \sin (e+f x)}}+\frac{\left (2 a^2 (c-d)\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d-d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{d f}\\ &=\frac{2 a^{3/2} (c-d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{d^{3/2} \sqrt{c+d} f}-\frac{2 a^2 \cos (e+f x)}{d f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 2.20081, size = 233, normalized size = 2.38 \[ \frac{(a (\sin (e+f x)+1))^{3/2} \left (2 \sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )-2 \sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+(c-d) \left (\log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-\log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )\right )\right )}{d^{3/2} f \sqrt{c+d} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[d]*Sqrt[c + d]*Cos[(e + f*x)/2] + (c - d)*(Log[-(Sec[(e + f*x)/4]^2*(c + d + Sqrt[d]*Sqrt[c + d]*Cos
[(e + f*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2]))] - Log[(c + d)*Sec[(e + f*x)/4]^2 + Sqrt[d]*Sqrt[c + d]
*(-1 + 2*Tan[(e + f*x)/4] + Tan[(e + f*x)/4]^2)]) + 2*Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*
x]))^(3/2))/(d^(3/2)*Sqrt[c + d]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [A]  time = 0.934, size = 137, normalized size = 1.4 \begin{align*} -2\,{\frac{a \left ( 1+\sin \left ( fx+e \right ) \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{d\sqrt{a \left ( c+d \right ) d}\cos \left ( fx+e \right ) \sqrt{a+a\sin \left ( fx+e \right ) }f} \left ( -{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) ac+a{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) d+\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d} \right ) } \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)),x)

[Out]

-2*a*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(-arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a*c+a
*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*d+(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2))/d/(a*
(c+d)*d)^(1/2)/cos(f*x+e)/(a+a*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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}{d \sin \left (f x + e\right ) + c}\,{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)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.34015, size = 1539, normalized size = 15.7 \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)),x, algorithm="fricas")

[Out]

[-1/2*((a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c - a*d)*sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*
x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2
+ d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(
f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e
) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*c
os(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e
)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(a*cos(f*x + e) - a*sin(f*x + e) + a)*sqrt(a*
sin(f*x + e) + a))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f), ((a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c
- a*d)*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(
-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(a*cos(f*x + e) - a*sin(f*x + e) + a)*sqrt(a*sin(f*x + e) + a))/(d*f*cos
(f*x + e) + d*f*sin(f*x + e) + d*f)]

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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))**(3/2)/(c+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [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))^(3/2)/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out