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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 172, normalized size of antiderivative = 1.00, 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 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 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 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]

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]

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 = 0.90, 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))

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 598, normalized size = 3.48 \[ \frac {2 \, {\left (2 \, {\left (a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 20 \, a c^{2} + 32 \, a c d - 12 \, a d^{2} - {\left (5 \, a c^{2} + 44 \, a c d - 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (25 \, a c^{2} + 14 \, a c d + 21 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (20 \, a c^{2} - 32 \, a c d + 12 \, a d^{2} - 2 \, {\left (a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (5 \, a c^{2} + 46 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{15 \, {\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{4} - 3 \, {\left (c^{4} d^{2} + 3 \, c^{3} d^{3} + 3 \, c^{2} d^{4} + c d^{5}\right )} f \cos \left (f x + e\right )^{3} - {\left (3 \, c^{5} d + 12 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 18 \, c^{2} d^{4} + 9 \, c d^{5} + 2 \, d^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (c^{6} + 3 \, c^{5} d + 6 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 9 \, c^{2} d^{4} + 3 \, c d^{5}\right )} f \cos \left (f x + e\right ) + {\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f - {\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{3} + {\left (3 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 12 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{2} - {\left (3 \, c^{5} d + 9 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 6 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right ) - {\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f\right )} \sin \left (f x + e\right )\right )}} \]

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))

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.36, size = 625, normalized size = 3.63 \[ -\frac {2 \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c +d \sin \left (f x +e \right )}\, \left (114 c^{3} \left (\cos ^{2}\left (f x +e \right )\right ) d^{2}+63 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) d^{5}-31 \left (\cos ^{2}\left (f x +e \right )\right ) c^{4} d -105 \left (\cos ^{4}\left (f x +e \right )\right ) c^{2} d^{3}+23 c \left (\cos ^{4}\left (f x +e \right )\right ) d^{4}-19 c \left (\cos ^{2}\left (f x +e \right )\right ) d^{4}-\left (\cos ^{6}\left (f x +e \right )\right ) c^{2} d^{3}-63 \left (\cos ^{4}\left (f x +e \right )\right ) c^{3} d^{2}-5 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c^{5}-13 \left (\cos ^{4}\left (f x +e \right )\right ) c^{4} d +18 \sin \left (f x +e \right ) \left (\cos ^{6}\left (f x +e \right )\right ) d^{5}-12 \left (\cos ^{6}\left (f x +e \right )\right ) c \,d^{4}+8 c \,d^{4}-80 c^{2} d^{3}-48 c^{3} d^{2}+56 c^{4} d -9 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right ) c \,d^{4}-90 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c^{3} d^{2}-146 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c^{2} d^{3}+15 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c \,d^{4}-24 d^{5} \sin \left (f x +e \right )+57 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right ) c^{2} d^{3}+3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c^{4} d +2 \sin \left (f x +e \right ) \left (\cos ^{6}\left (f x +e \right )\right ) c \,d^{4}+9 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right ) c^{3} d^{2}+80 c^{2} d^{3} \sin \left (f x +e \right )-8 c \,d^{4} \sin \left (f x +e \right )-56 \sin \left (f x +e \right ) c^{4} d +48 \sin \left (f x +e \right ) c^{3} d^{2}+24 d^{5}-57 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right ) d^{5}+186 \left (\cos ^{2}\left (f x +e \right )\right ) c^{2} d^{3}-15 \left (\cos ^{2}\left (f x +e \right )\right ) c^{5}-27 \left (\cos ^{6}\left (f x +e \right )\right ) d^{5}-75 \left (\cos ^{2}\left (f x +e \right )\right ) d^{5}+78 \left (\cos ^{4}\left (f x +e \right )\right ) d^{5}+40 c^{5}-40 c^{5} \sin \left (f x +e \right )\right )}{15 f \cos \left (f x +e \right )^{3} \left (\left (\cos ^{2}\left (f x +e \right )\right ) d^{2}+c^{2}-d^{2}\right )^{3} \left (c +d \right )^{3}} \]

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*(a*(1+sin(f*x+e)))^(3/2)*(c+d*sin(f*x+e))^(1/2)*(-57*sin(f*x+e)*cos(f*x+e)^4*d^5-9*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+8*c*d^4-80*c^2*d^3-48*c^3*d^2+56*c^4*d+186*cos(f*x+e)^2*c^2*d^3+18*sin(f*x+e)*cos(f*x+e)^6*d^5-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+7
8*cos(f*x+e)^4*d^5-75*cos(f*x+e)^2*d^5-27*cos(f*x+e)^6*d^5-24*d^5*sin(f*x+e)+57*sin(f*x+e)*cos(f*x+e)^4*c^2*d^
3+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+23*c*cos(f
*x+e)^4*d^4+114*c^3*cos(f*x+e)^2*d^2-19*c*cos(f*x+e)^2*d^4+80*c^2*d^3*sin(f*x+e)-8*c*d^4*sin(f*x+e)-105*cos(f*
x+e)^4*c^2*d^3+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+
24*d^5-15*cos(f*x+e)^2*c^5+40*c^5-40*c^5*sin(f*x+e))/cos(f*x+e)^3/(cos(f*x+e)^2*d^2+c^2-d^2)^3/(c+d)^3

________________________________________________________________________________________

maxima [B]  time = 1.22, size = 505, normalized size = 2.94 \[ -\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} \]

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)

________________________________________________________________________________________

mupad [B]  time = 17.48, size = 541, normalized size = 3.15 \[ \frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}\,\left (\frac {8\,a\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (c+9\,d\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{15\,d^2\,f\,{\left (c+d\right )}^3}-\frac {8\,a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\left (9\,c^2-4\,c\,d+3\,d^2\right )}{3\,d^3\,f\,{\left (c+d\right )}^3}+\frac {8\,a\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\left (c^2\,9{}\mathrm {i}-c\,d\,4{}\mathrm {i}+d^2\,3{}\mathrm {i}\right )}{3\,d^3\,f\,{\left (c+d\right )}^3}-\frac {8\,a\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (c\,1{}\mathrm {i}+d\,9{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{15\,d^2\,f\,{\left (c+d\right )}^3}+\frac {8\,a\,c\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\left (c\,1{}\mathrm {i}+d\,9{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,d^3\,f\,{\left (c+d\right )}^3}-\frac {8\,a\,c\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (c+9\,d\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,d^3\,f\,{\left (c+d\right )}^3}\right )}{{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}+\frac {{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^3}{{\left (c+d\right )}^3}-\frac {3\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\left (4\,c^2+2\,c\,d+d^2\right )}{d^2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (6\,c+d\right )}{d}+\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (8\,c^3+12\,c^2\,d+12\,c\,d^2+3\,d^3\right )}{d^3}+\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (c\,6{}\mathrm {i}+d\,1{}\mathrm {i}\right )}{d}-\frac {3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^3\,\left (4\,c^2+2\,c\,d+d^2\right )}{d^2\,{\left (c+d\right )}^3}+\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^3\,\left (8\,c^3+12\,c^2\,d+12\,c\,d^2+3\,d^3\right )}{d^3\,{\left (c+d\right )}^3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*sin(e + f*x))^(1/2)*((8*a*exp(e*6i + f*x*6i)*(c + 9*d)*(a + a*sin(e + f*x))^(1/2))/(15*d^2*f*(c + d)^3
) - (8*a*exp(e*4i + f*x*4i)*(a + a*sin(e + f*x))^(1/2)*(9*c^2 - 4*c*d + 3*d^2))/(3*d^3*f*(c + d)^3) + (8*a*exp
(e*3i + f*x*3i)*(a + a*sin(e + f*x))^(1/2)*(c^2*9i - c*d*4i + d^2*3i))/(3*d^3*f*(c + d)^3) - (8*a*exp(e*1i + f
*x*1i)*(c*1i + d*9i)*(a + a*sin(e + f*x))^(1/2))/(15*d^2*f*(c + d)^3) + (8*a*c*exp(e*5i + f*x*5i)*(c*1i + d*9i
)*(a + a*sin(e + f*x))^(1/2))/(3*d^3*f*(c + d)^3) - (8*a*c*exp(e*2i + f*x*2i)*(c + 9*d)*(a + a*sin(e + f*x))^(
1/2))/(3*d^3*f*(c + d)^3)))/(exp(e*7i + f*x*7i) + (c*1i + d*1i)^3/(c + d)^3 - (3*exp(e*5i + f*x*5i)*(2*c*d + 4
*c^2 + d^2))/d^2 - (exp(e*1i + f*x*1i)*(6*c + d))/d + (exp(e*3i + f*x*3i)*(12*c*d^2 + 12*c^2*d + 8*c^3 + 3*d^3
))/d^3 + (exp(e*6i + f*x*6i)*(c*6i + d*1i))/d - (3*exp(e*2i + f*x*2i)*(c*1i + d*1i)^3*(2*c*d + 4*c^2 + d^2))/(
d^2*(c + d)^3) + (exp(e*4i + f*x*4i)*(c*1i + d*1i)^3*(12*c*d^2 + 12*c^2*d + 8*c^3 + 3*d^3))/(d^3*(c + d)^3))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________