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

Optimal. Leaf size=115 \[ \frac{2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{2 a^2 (c+5 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.214542, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2762, 21, 2771} \[ \frac{2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{2 a^2 (c+5 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (2*a^2*(c +
5*d)*Cos[e + f*x])/(3*d*(c + d)^2*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 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*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]

Rubi steps

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

Mathematica [A]  time = 0.586816, size = 104, normalized size = 0.9 \[ -\frac{2 a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) ((c+5 d) \sin (e+f x)+5 c+d)}{3 f (c+d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.208, size = 345, normalized size = 3. \begin{align*} -{\frac{2\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}c{d}^{2}+10\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{3}-4\,{c}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{4}d-14\,c \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{2}-18\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{3}-2\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{3}+2\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{2}d-22\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}c{d}^{2}-26\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{3}-6\,{c}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}-10\,{c}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}d+30\,c \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}+34\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{3}-16\,{c}^{3}\sin \left ( fx+e \right ) -16\,{c}^{2}d\sin \left ( fx+e \right ) +16\,\sin \left ( fx+e \right ){d}^{2}c+16\,{d}^{3}\sin \left ( fx+e \right ) +16\,{c}^{3}+16\,{c}^{2}d-16\,c{d}^{2}-16\,{d}^{3}}{3\,f \left ( c+d \right ) ^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}+{c}^{2}-{d}^{2} \right ) ^{2}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}\sqrt{c+d\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))^(5/2),x)

[Out]

-2/3/f/(c+d)^2*(a*(1+sin(f*x+e)))^(3/2)*(c+d*sin(f*x+e))^(1/2)*(sin(f*x+e)*cos(f*x+e)^4*c*d^2+5*sin(f*x+e)*cos
(f*x+e)^4*d^3-2*c^2*cos(f*x+e)^4*d-7*c*cos(f*x+e)^4*d^2-9*cos(f*x+e)^4*d^3-sin(f*x+e)*cos(f*x+e)^2*c^3+sin(f*x
+e)*cos(f*x+e)^2*c^2*d-11*sin(f*x+e)*cos(f*x+e)^2*c*d^2-13*sin(f*x+e)*cos(f*x+e)^2*d^3-3*c^3*cos(f*x+e)^2-5*c^
2*cos(f*x+e)^2*d+15*c*cos(f*x+e)^2*d^2+17*cos(f*x+e)^2*d^3-8*c^3*sin(f*x+e)-8*c^2*d*sin(f*x+e)+8*sin(f*x+e)*d^
2*c+8*d^3*sin(f*x+e)+8*c^3+8*c^2*d-8*c*d^2-8*d^3)/cos(f*x+e)^3/(cos(f*x+e)^2*d^2+c^2-d^2)^2

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Maxima [B]  time = 1.7358, size = 414, normalized size = 3.6 \begin{align*} -\frac{2 \,{\left ({\left (5 \, c^{2} + c d\right )} a^{\frac{3}{2}} - \frac{{\left (3 \, c^{2} - 19 \, c d - 2 \, d^{2}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \,{\left (4 \, c^{2} - 7 \, c d + 9 \, d^{2}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{2 \,{\left (4 \, c^{2} - 7 \, c d + 9 \, d^{2}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{{\left (3 \, c^{2} - 19 \, c d - 2 \, d^{2}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{{\left (5 \, c^{2} + c d\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{3 \,{\left (c^{2} + 2 \, c d + d^{2} + \frac{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (c + \frac{2 \, d \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{\frac{5}{2}} f} \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))^(5/2),x, algorithm="maxima")

[Out]

-2/3*((5*c^2 + c*d)*a^(3/2) - (3*c^2 - 19*c*d - 2*d^2)*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*(4*c^2 - 7*
c*d + 9*d^2)*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*(4*c^2 - 7*c*d + 9*d^2)*a^(3/2)*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + (3*c^2 - 19*c*d - 2*d^2)*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (5*c^2 + c*d)*a^(3
/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/((c^2 + 2*c*d + d^2 + (c^2
+ 2*c*d + d^2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^
2/(cos(f*x + e) + 1)^2)^(5/2)*f)

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Fricas [B]  time = 2.69004, size = 752, normalized size = 6.54 \begin{align*} \frac{2 \,{\left ({\left (a c + 5 \, a d\right )} \cos \left (f x + e\right )^{2} + 4 \, a c - 4 \, a d +{\left (5 \, a c + a d\right )} \cos \left (f x + e\right ) -{\left (4 \, a c - 4 \, a d -{\left (a c + 5 \, a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{d \sin \left (f x + e\right ) + c}}{3 \,{\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{3} +{\left (2 \, c^{3} d + 5 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} -{\left (c^{4} + 2 \, c^{3} d + 2 \, c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right ) -{\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f +{\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - 2 \,{\left (c^{3} d + 2 \, c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) -{\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}} \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))^(5/2),x, algorithm="fricas")

[Out]

2/3*((a*c + 5*a*d)*cos(f*x + e)^2 + 4*a*c - 4*a*d + (5*a*c + a*d)*cos(f*x + e) - (4*a*c - 4*a*d - (a*c + 5*a*d
)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/((c^2*d^2 + 2*c*d^3 + d^4)*f*c
os(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3
+ d^4)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x +
e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*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))**(3/2)/(c+d*sin(f*x+e))**(5/2),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{3}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{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))^(5/2),x, algorithm="giac")

[Out]

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