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

Optimal. Leaf size=142 \[ \frac{2 a^3 (3 c-7 d) \cos (e+f x)}{3 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{2 a^{5/2} (c-d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{5/2} f \sqrt{c+d}}-\frac{2 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 d f} \]

[Out]

(-2*a^(5/2)*(c - d)^2*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(5/2)
*Sqrt[c + d]*f) + (2*a^3*(3*c - 7*d)*Cos[e + f*x])/(3*d^2*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*Cos[e + f*x]*Sq
rt[a + a*Sin[e + f*x]])/(3*d*f)

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Rubi [A]  time = 0.411399, antiderivative size = 142, 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, 2981, 2773, 208} \[ \frac{2 a^3 (3 c-7 d) \cos (e+f x)}{3 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{2 a^{5/2} (c-d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{5/2} f \sqrt{c+d}}-\frac{2 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^(5/2)*(c - d)^2*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(5/2)
*Sqrt[c + d]*f) + (2*a^3*(3*c - 7*d)*Cos[e + f*x])/(3*d^2*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*Cos[e + f*x]*Sq
rt[a + a*Sin[e + f*x]])/(3*d*f)

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 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 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))^{5/2}}{c+d \sin (e+f x)} \, dx &=-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d f}+\frac{2 \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a^2 (c+3 d)-\frac{1}{2} a^2 (3 c-7 d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{3 d}\\ &=\frac{2 a^3 (3 c-7 d) \cos (e+f x)}{3 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d f}+\frac{\left (a^2 (c-d)^2\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{d^2}\\ &=\frac{2 a^3 (3 c-7 d) \cos (e+f x)}{3 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d f}-\frac{\left (2 a^3 (c-d)^2\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^2 f}\\ &=-\frac{2 a^{5/2} (c-d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{d^{5/2} \sqrt{c+d} f}+\frac{2 a^3 (3 c-7 d) \cos (e+f x)}{3 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 d f}\\ \end{align*}

Mathematica [B]  time = 3.59531, size = 330, normalized size = 2.32 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (6 \sqrt{d} (5 d-2 c) \sin \left (\frac{1}{2} (e+f x)\right )+6 \sqrt{d} (2 c-5 d) \cos \left (\frac{1}{2} (e+f x)\right )+\frac{3 (c-d)^2 \left (2 \log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt{c+d}}-\frac{3 (c-d)^2 \left (2 \log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt{c+d}}-2 d^{3/2} \sin \left (\frac{3}{2} (e+f x)\right )-2 d^{3/2} \cos \left (\frac{3}{2} (e+f x)\right )\right )}{6 d^{5/2} f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(6*(2*c - 5*d)*Sqrt[d]*Cos[(e + f*x)/2] - 2*d^(3/2)*Cos[(3*(e + f*x))/2] - (3*(c
- d)^2*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*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]))]))/Sqrt[c + d] + (3*(c - d)^2*(e + f*x - 2*Log[Sec[(e + f*x
)/4]^2] + 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
)]))/Sqrt[c + d] + 6*Sqrt[d]*(-2*c + 5*d)*Sin[(e + f*x)/2] - 2*d^(3/2)*Sin[(3*(e + f*x))/2]))/(6*d^(5/2)*f*(Co
s[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [A]  time = 0.804, size = 228, normalized size = 1.6 \begin{align*}{\frac{2\,a \left ( 1+\sin \left ( fx+e \right ) \right ) }{3\,{d}^{2}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( \left ( -a \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}\sqrt{a \left ( c+d \right ) d}d-3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ){a}^{2}{c}^{2}+6\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ){a}^{2}cd-3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ){a}^{2}{d}^{2}+3\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d}ac-9\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d}ad \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))^(5/2)/(c+d*sin(f*x+e)),x)

[Out]

2/3*a*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*((-a*(-1+sin(f*x+e)))^(3/2)*(a*(c+d)*d)^(1/2)*d-3*arctanh((-a*
(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^2+6*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*
a^2*c*d-3*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*d^2+3*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d
)*d)^(1/2)*a*c-9*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a*d)/d^2/(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{5}{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))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.52201, size = 1987, normalized size = 13.99 \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)),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e) + (a^2*c^2 - 2*a^2*c*d +
a^2*d^2)*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^2*d*cos(f*x + e)^2 - 3*a^2*c + 7*a^2*d - (3*a^2*c - 8*a^2*d)*cos(f*x + e) + (a^2*d*cos(f*x
+ e) + 3*a^2*c - 7*a^2*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^2*f*cos(f*x + e) + d^2*f*sin(f*x + e) + d
^2*f), -1/3*(3*(a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e) + (a^2*c^2 - 2*a^
2*c*d + a^2*d^2)*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^2*d*cos(f*x + e)^2 - 3*a^2*c + 7*a^2*d - (3*a^2*c - 8*a^2*d
)*cos(f*x + e) + (a^2*d*cos(f*x + e) + 3*a^2*c - 7*a^2*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^2*f*cos(f
*x + e) + d^2*f*sin(f*x + e) + d^2*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))**(5/2)/(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))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out