3.577 \(\int \frac{(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac{4 a^2 (c+9 d) \cos (e+f x)}{15 d f (c+d)^3 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^2 (c+9 d) \cos (e+f x)}{15 d f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}+\frac{2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}} \]

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(5*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)) - (2*a^2*(c +
9*d)*Cos[e + f*x])/(15*d*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c + 9*d)*
Cos[e + f*x])/(15*d*(c + d)^3*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.324019, antiderivative size = 172, 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, 2772, 2771} \[ -\frac{4 a^2 (c+9 d) \cos (e+f x)}{15 d f (c+d)^3 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^2 (c+9 d) \cos (e+f x)}{15 d f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}+\frac{2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(5*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)) - (2*a^2*(c +
9*d)*Cos[e + f*x])/(15*d*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c + 9*d)*
Cos[e + f*x])/(15*d*(c + d)^3*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 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

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))^{7/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{(2 a) \int \frac{-\frac{1}{2} a (c+9 d)-\frac{1}{2} a (c+9 d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}+\frac{(a (c+9 d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{2 a^2 (c+9 d) \cos (e+f x)}{15 d (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{(2 a (c+9 d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 d (c+d)^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{2 a^2 (c+9 d) \cos (e+f x)}{15 d (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{4 a^2 (c+9 d) \cos (e+f x)}{15 d (c+d)^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.888506, size = 140, normalized size = 0.81 \[ -\frac{2 a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\left (5 c^2+46 c d+9 d^2\right ) \sin (e+f x)+25 c^2-d (c+9 d) \cos (2 (e+f x))+13 c d+12 d^2\right )}{15 f (c+d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(-2*a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(25*c^2 + 13*c*d + 12*d^2 - d*(c + 9*d)
*Cos[2*(e + f*x)] + (5*c^2 + 46*c*d + 9*d^2)*Sin[e + f*x]))/(15*(c + d)^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2])*(c + d*Sin[e + f*x])^(5/2))

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Maple [B]  time = 0.234, size = 625, normalized size = 3.6 \begin{align*} \text{result 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))^(7/2),x)

[Out]

-2/15/f/(c+d)^3*(a*(1+sin(f*x+e)))^(3/2)*(c+d*sin(f*x+e))^(1/2)*(23*c*cos(f*x+e)^4*d^4+24*d^5+186*cos(f*x+e)^2
*c^2*d^3+56*c^4*d-80*c^2*d^3+8*c*d^4+78*cos(f*x+e)^4*d^5-75*cos(f*x+e)^2*d^5-27*cos(f*x+e)^6*d^5-40*c^5*sin(f*
x+e)+80*c^2*d^3*sin(f*x+e)-8*c*d^4*sin(f*x+e)-cos(f*x+e)^6*c^2*d^3-12*cos(f*x+e)^6*c*d^4-13*cos(f*x+e)^4*c^4*d
-63*cos(f*x+e)^4*c^3*d^2-5*sin(f*x+e)*cos(f*x+e)^2*c^5-24*d^5*sin(f*x+e)+18*sin(f*x+e)*cos(f*x+e)^6*d^5-48*c^3
*d^2+3*sin(f*x+e)*cos(f*x+e)^2*c^4*d+2*sin(f*x+e)*cos(f*x+e)^6*c*d^4+9*sin(f*x+e)*cos(f*x+e)^4*c^3*d^2+57*sin(
f*x+e)*cos(f*x+e)^4*c^2*d^3-9*sin(f*x+e)*cos(f*x+e)^4*c*d^4-90*sin(f*x+e)*cos(f*x+e)^2*c^3*d^2-146*sin(f*x+e)*
cos(f*x+e)^2*c^2*d^3+15*sin(f*x+e)*cos(f*x+e)^2*c*d^4+114*c^3*cos(f*x+e)^2*d^2-19*c*d^4*cos(f*x+e)^2-57*sin(f*
x+e)*cos(f*x+e)^4*d^5+63*sin(f*x+e)*cos(f*x+e)^2*d^5-31*cos(f*x+e)^2*c^4*d-56*sin(f*x+e)*c^4*d+48*sin(f*x+e)*c
^3*d^2-105*cos(f*x+e)^4*c^2*d^3+40*c^5-15*cos(f*x+e)^2*c^5)/cos(f*x+e)^3/(cos(f*x+e)^2*d^2+c^2-d^2)^3

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Maxima [B]  time = 2.08734, size = 682, normalized size = 3.97 \begin{align*} -\frac{2 \,{\left ({\left (25 \, c^{3} + 12 \, c^{2} d + 3 \, c d^{2}\right )} a^{\frac{3}{2}} - \frac{{\left (15 \, c^{3} - 130 \, c^{2} d - 39 \, c d^{2} - 6 \, d^{3}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{{\left (65 \, c^{3} - 78 \, c^{2} d + 223 \, c d^{2} + 30 \, d^{3}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{5 \,{\left (11 \, c^{3} - 44 \, c^{2} d + 33 \, c d^{2} - 24 \, d^{3}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \,{\left (11 \, c^{3} - 44 \, c^{2} d + 33 \, c d^{2} - 24 \, d^{3}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{{\left (65 \, c^{3} - 78 \, c^{2} d + 223 \, c d^{2} + 30 \, d^{3}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{{\left (15 \, c^{3} - 130 \, c^{2} d - 39 \, c d^{2} - 6 \, d^{3}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{{\left (25 \, c^{3} + 12 \, c^{2} d + 3 \, c d^{2}\right )} a^{\frac{3}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{2}}{15 \,{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3} + \frac{2 \,{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

-2/15*((25*c^3 + 12*c^2*d + 3*c*d^2)*a^(3/2) - (15*c^3 - 130*c^2*d - 39*c*d^2 - 6*d^3)*a^(3/2)*sin(f*x + e)/(c
os(f*x + e) + 1) + (65*c^3 - 78*c^2*d + 223*c*d^2 + 30*d^3)*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*(1
1*c^3 - 44*c^2*d + 33*c*d^2 - 24*d^3)*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*(11*c^3 - 44*c^2*d + 33*
c*d^2 - 24*d^3)*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (65*c^3 - 78*c^2*d + 223*c*d^2 + 30*d^3)*a^(3/2)
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + (15*c^3 - 130*c^2*d - 39*c*d^2 - 6*d^3)*a^(3/2)*sin(f*x + e)^6/(cos(f*x
+ e) + 1)^6 - (25*c^3 + 12*c^2*d + 3*c*d^2)*a^(3/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*(sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 1)^2/((c^3 + 3*c^2*d + 3*c*d^2 + d^3 + 2*(c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sin(f*x + e)^2/(co
s(f*x + e) + 1)^2 + (c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(c + 2*d*sin(f*x + e)
/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)^(7/2)*f)

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Fricas [B]  time = 3.32449, size = 1369, normalized size = 7.96 \begin{align*} \frac{2 \,{\left (2 \,{\left (a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 20 \, a c^{2} + 32 \, a c d - 12 \, a d^{2} -{\left (5 \, a c^{2} + 44 \, a c d - 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (25 \, a c^{2} + 14 \, a c d + 21 \, a d^{2}\right )} \cos \left (f x + e\right ) +{\left (20 \, a c^{2} - 32 \, a c d + 12 \, a d^{2} - 2 \,{\left (a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (5 \, a c^{2} + 46 \, a c d + 9 \, a d^{2}\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}}{15 \,{\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{4} - 3 \,{\left (c^{4} d^{2} + 3 \, c^{3} d^{3} + 3 \, c^{2} d^{4} + c d^{5}\right )} f \cos \left (f x + e\right )^{3} -{\left (3 \, c^{5} d + 12 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 18 \, c^{2} d^{4} + 9 \, c d^{5} + 2 \, d^{6}\right )} f \cos \left (f x + e\right )^{2} +{\left (c^{6} + 3 \, c^{5} d + 6 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 9 \, c^{2} d^{4} + 3 \, c d^{5}\right )} f \cos \left (f x + e\right ) +{\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f -{\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{3} +{\left (3 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 12 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{2} -{\left (3 \, c^{5} d + 9 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 6 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right ) -{\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\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))^(7/2),x, algorithm="fricas")

[Out]

2/15*(2*(a*c*d + 9*a*d^2)*cos(f*x + e)^3 - 20*a*c^2 + 32*a*c*d - 12*a*d^2 - (5*a*c^2 + 44*a*c*d - 9*a*d^2)*cos
(f*x + e)^2 - (25*a*c^2 + 14*a*c*d + 21*a*d^2)*cos(f*x + e) + (20*a*c^2 - 32*a*c*d + 12*a*d^2 - 2*(a*c*d + 9*a
*d^2)*cos(f*x + e)^2 - (5*a*c^2 + 46*a*c*d + 9*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqr
t(d*sin(f*x + e) + c)/((c^3*d^3 + 3*c^2*d^4 + 3*c*d^5 + d^6)*f*cos(f*x + e)^4 - 3*(c^4*d^2 + 3*c^3*d^3 + 3*c^2
*d^4 + c*d^5)*f*cos(f*x + e)^3 - (3*c^5*d + 12*c^4*d^2 + 20*c^3*d^3 + 18*c^2*d^4 + 9*c*d^5 + 2*d^6)*f*cos(f*x
+ e)^2 + (c^6 + 3*c^5*d + 6*c^4*d^2 + 10*c^3*d^3 + 9*c^2*d^4 + 3*c*d^5)*f*cos(f*x + e) + (c^6 + 6*c^5*d + 15*c
^4*d^2 + 20*c^3*d^3 + 15*c^2*d^4 + 6*c*d^5 + d^6)*f - ((c^3*d^3 + 3*c^2*d^4 + 3*c*d^5 + d^6)*f*cos(f*x + e)^3
+ (3*c^4*d^2 + 10*c^3*d^3 + 12*c^2*d^4 + 6*c*d^5 + d^6)*f*cos(f*x + e)^2 - (3*c^5*d + 9*c^4*d^2 + 10*c^3*d^3 +
6*c^2*d^4 + 3*c*d^5 + d^6)*f*cos(f*x + e) - (c^6 + 6*c^5*d + 15*c^4*d^2 + 20*c^3*d^3 + 15*c^2*d^4 + 6*c*d^5 +
d^6)*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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)/(d*sin(f*x + e) + c)^(7/2), x)