\(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx\) [585]

Optimal result

Integrand size = 29, antiderivative size = 172 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac {2 a^2 (c+9 d) \cos (e+f x)}{15 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+9 d) \cos (e+f x)}{15 d (c+d)^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {2 a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (25 c^2+13 c d+12 d^2-d (c+9 d) \cos (2 (e+f x))+\left (5 c^2+46 c d+9 d^2\right ) \sin (e+f x)\right )}{15 (c+d)^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{5/2}} \] Input:

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

Output:

(-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))
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3241, 27, 2011, 3042, 3251, 3042, 3250}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}-\frac {2 a \int -\frac {a (c+9 d)+a \sin (e+f x) (c+9 d)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \int \frac {a (c+9 d)+a \sin (e+f x) (c+9 d)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {a (c+9 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (c+9 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {a (c+9 d) \left (\frac {2 \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{3/2}}dx}{3 (c+d)}-\frac {2 a \cos (e+f x)}{3 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\right )}{5 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (c+9 d) \left (\frac {2 \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{3/2}}dx}{3 (c+d)}-\frac {2 a \cos (e+f x)}{3 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\right )}{5 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3250

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}+\frac {a (c+9 d) \left (-\frac {4 a \cos (e+f x)}{3 f (c+d)^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 a \cos (e+f x)}{3 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\right )}{5 d (c+d)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] + Simp[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 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[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 3251

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] + Sim 
p[(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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(458\) vs. \(2(154)=308\).

Time = 22.69 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.67

method	result	size
default	\(\frac {2 \left (\left (\left (-10 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+25\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-10 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{3}+\left (\left (-28 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-114 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+12\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+28 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-170 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+130 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{2} d +\left (\left (16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-272 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+214 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+224 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-282 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+39 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2} c +\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\left (-144 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-144 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+108 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+36 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+6\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-6 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{3}\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, a}{15 f \left (c^{3}+3 c^{2} d +3 d^{2} c +d^{3}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (c +2 d \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{\frac {7}{2}}}\)	\(459\)

Input:

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

Output:

2/15/f/(c^3+3*c^2*d+3*c*d^2+d^3)*(((-10*cos(1/2*f*x+1/2*e)^2+25)*sin(1/2*f 
*x+1/2*e)-10*cos(1/2*f*x+1/2*e)^3-15*cos(1/2*f*x+1/2*e))*c^3+((-28*cos(1/2 
*f*x+1/2*e)^4-114*cos(1/2*f*x+1/2*e)^2+12)*sin(1/2*f*x+1/2*e)+28*cos(1/2*f 
*x+1/2*e)^5-170*cos(1/2*f*x+1/2*e)^3+130*cos(1/2*f*x+1/2*e))*c^2*d+((16*co 
s(1/2*f*x+1/2*e)^6-272*cos(1/2*f*x+1/2*e)^4+214*cos(1/2*f*x+1/2*e)^2+3)*si 
n(1/2*f*x+1/2*e)+16*cos(1/2*f*x+1/2*e)^7+224*cos(1/2*f*x+1/2*e)^5-282*cos( 
1/2*f*x+1/2*e)^3+39*cos(1/2*f*x+1/2*e))*d^2*c+cos(1/2*f*x+1/2*e)*((-144*co 
s(1/2*f*x+1/2*e)^4-144*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)^3+108*cos(1/2 
*f*x+1/2*e)^2+36*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+6)*sin(1/2*f*x+1/2* 
e)^2-6*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))*d^3)*((2*sin(1/2*f*x+1/2*e)* 
cos(1/2*f*x+1/2*e)+1)*a)^(1/2)*a/(cos(1/2*f*x+1/2*e)+sin(1/2*f*x+1/2*e))/( 
c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))^(7/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 598 vs. \(2 (154) = 308\).

Time = 0.14 (sec) , antiderivative size = 598, normalized size of antiderivative = 3.48 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\frac {2 \, {\left (2 \, {\left (a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 20 \, a c^{2} + 32 \, a c d - 12 \, a d^{2} - {\left (5 \, a c^{2} + 44 \, a c d - 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (25 \, a c^{2} + 14 \, a c d + 21 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (20 \, a c^{2} - 32 \, a c d + 12 \, a d^{2} - 2 \, {\left (a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (5 \, a c^{2} + 46 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{15 \, {\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{4} - 3 \, {\left (c^{4} d^{2} + 3 \, c^{3} d^{3} + 3 \, c^{2} d^{4} + c d^{5}\right )} f \cos \left (f x + e\right )^{3} - {\left (3 \, c^{5} d + 12 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 18 \, c^{2} d^{4} + 9 \, c d^{5} + 2 \, d^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (c^{6} + 3 \, c^{5} d + 6 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 9 \, c^{2} d^{4} + 3 \, c d^{5}\right )} f \cos \left (f x + e\right ) + {\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f - {\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{3} + {\left (3 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 12 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{2} - {\left (3 \, c^{5} d + 9 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 6 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right ) - {\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="fric 
as")
 

Output:

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 + 2 
1*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)*sqrt(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))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (154) = 308\).

Time = 0.21 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.94 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=-\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} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="maxi 
ma")
                                                                                    
                                                                                    
 

Output:

-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)/(cos(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*(11* 
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/(cos(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)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 27.64 (sec) , antiderivative size = 541, normalized size of antiderivative = 3.15 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

((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))
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\sqrt {a}\, a \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4} d^{4}+4 \sin \left (f x +e \right )^{3} c \,d^{3}+6 \sin \left (f x +e \right )^{2} c^{2} d^{2}+4 \sin \left (f x +e \right ) c^{3} d +c^{4}}d x +\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4} d^{4}+4 \sin \left (f x +e \right )^{3} c \,d^{3}+6 \sin \left (f x +e \right )^{2} c^{2} d^{2}+4 \sin \left (f x +e \right ) c^{3} d +c^{4}}d x \right ) \] Input:

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

Output:

sqrt(a)*a*(int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f* 
x))/(sin(e + f*x)**4*d**4 + 4*sin(e + f*x)**3*c*d**3 + 6*sin(e + f*x)**2*c 
**2*d**2 + 4*sin(e + f*x)*c**3*d + c**4),x) + int((sqrt(sin(e + f*x)*d + c 
)*sqrt(sin(e + f*x) + 1))/(sin(e + f*x)**4*d**4 + 4*sin(e + f*x)**3*c*d**3 
 + 6*sin(e + f*x)**2*c**2*d**2 + 4*sin(e + f*x)*c**3*d + c**4),x))