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

Optimal. Leaf size=119 \[ \frac{a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{a^{3/2} (c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.192916, antiderivative size = 119, 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 = {2762, 21, 2773, 208} \[ \frac{a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{a^{3/2} (c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*(c + 3*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(3/2)*
(c + d)^(3/2)*f)) + (a^2*(c - d)*Cos[e + f*x])/(d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(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 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))^2} \, dx &=\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{a \int \frac{-\frac{1}{2} a (c+3 d)-\frac{1}{2} a (c+3 d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{d (c+d)}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac{(a (c+3 d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 d (c+d)}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{\left (a^2 (c+3 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 (c+d) f}\\ &=-\frac{a^{3/2} (c+3 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} (c+d)^{3/2} f}+\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [B]  time = 2.35363, size = 268, normalized size = 2.25 \[ -\frac{(a (\sin (e+f x)+1))^{3/2} \left (2 \sqrt{d} (c-d) \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )-2 \sqrt{d} (c-d) \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+(c+3 d) (c+d \sin (e+f x)) \left (\log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-\log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )\right )\right )}{2 d^{3/2} f (c+d)^{3/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.197, size = 233, normalized size = 2. \begin{align*}{\frac{a \left ( 1+\sin \left ( fx+e \right ) \right ) }{d \left ( c+d \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( -\sin \left ( fx+e \right ){\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) ad \left ( c+3\,d \right ) -{\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) a{c}^{2}-3\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) acd+\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}c-\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}d \right ){\frac{1}{\sqrt{a \left ( c+d \right ) d}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \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))^2,x)

[Out]

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

[Out]

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

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Fricas [B]  time = 2.60626, size = 2241, normalized size = 18.83 \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))^2,x, algorithm="fricas")

[Out]

[-1/4*((a*c^2 + 4*a*c*d + 3*a*d^2 - (a*c*d + 3*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 3*a*c*d)*cos(f*x + e) + (a*c^2
+ 4*a*c*d + 3*a*d^2 + (a*c*d + 3*a*d^2)*cos(f*x + e))*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*cos(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*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))/((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 + 4*a*c*d + 3*a*d^2 - (a*c*d + 3*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 3*a*c*d)*cos(f*x + e) + (a*c^2 + 4*a*c*d
+ 3*a*d^2 + (a*c*d + 3*a*d^2)*cos(f*x + e))*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*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))/((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(-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))**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError