\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx\) [593]

Optimal result

Integrand size = 29, antiderivative size = 189 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac {2 a^3 (c-d) (3 c+11 d) \cos (e+f x)}{15 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {2 a^3 \left (3 c^2+14 c d+43 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \] Output:

2/5*a^2*(c-d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d/(c+d)/f/(c+d*sin(f*x+e)) 
^(5/2)+2/15*a^3*(c-d)*(3*c+11*d)*cos(f*x+e)/d^2/(c+d)^2/f/(a+a*sin(f*x+e)) 
^(1/2)/(c+d*sin(f*x+e))^(3/2)-2/15*a^3*(3*c^2+14*c*d+43*d^2)*cos(f*x+e)/d^ 
2/(c+d)^3/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 4.80 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.80 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (89 c^2+42 c d+49 d^2-\left (3 c^2+14 c d+43 d^2\right ) \cos (2 (e+f x))+4 \left (7 c^2+46 c d+7 d^2\right ) \sin (e+f x)\right )}{15 (c+d)^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{5/2}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(7/2),x]
 

Output:

-1/15*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x]) 
]*(89*c^2 + 42*c*d + 49*d^2 - (3*c^2 + 14*c*d + 43*d^2)*Cos[2*(e + f*x)] + 
 4*(7*c^2 + 46*c*d + 7*d^2)*Sin[e + f*x]))/((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 = 0.89 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3241, 27, 3042, 3459, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c+d \sin (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3241

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}}-\frac {2 a \int \frac {\sqrt {\sin (e+f x) a+a} (a (c-11 d)-a (3 c+7 d) \sin (e+f x))}{2 (c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a} (a (c-11 d)-a (3 c+7 d) \sin (e+f x))}{(c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a} (a (c-11 d)-a (3 c+7 d) \sin (e+f x))}{(c+d \sin (e+f x))^{5/2}}dx}{5 d (c+d)}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}}-\frac {a \left (-\frac {a \left (3 c^2+14 c d+43 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}-\frac {2 a^2 (c-d) (3 c+11 d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\right )}{5 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}}-\frac {a \left (-\frac {a \left (3 c^2+14 c d+43 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}-\frac {2 a^2 (c-d) (3 c+11 d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\right )}{5 d (c+d)}\)

\(\Big \downarrow \) 3250

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}}-\frac {a \left (\frac {2 a^2 \left (3 c^2+14 c d+43 d^2\right ) \cos (e+f x)}{3 d f (c+d)^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 (c-d) (3 c+11 d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\right )}{5 d (c+d)}\)

Input:

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(7/2),x]
 

Output:

(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(5*d*(c + d)*f*(c + 
d*Sin[e + f*x])^(5/2)) - (a*((-2*a^2*(c - d)*(3*c + 11*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 
*(3*c^2 + 14*c*d + 43*d^2)*Cos[e + f*x])/(3*d*(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 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 3459

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(512\) vs. \(2(171)=342\).

Time = 24.11 (sec) , antiderivative size = 513, normalized size of antiderivative = 2.71

method	result	size
default	\(\frac {2 \left (\left (\left (-12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+43\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-40 \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 (24 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-136 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-158 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+14\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+24 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+64 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-358 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+256 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{2} d +\left (\left (112 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-652 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+484 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+112 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+316 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-484 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+53 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c \,d^{2}+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\left (-344 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-344 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+288 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+56 \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^{2}}{15 f \left (c^{3}+3 c^{2} d +3 c \,d^{2}+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}}}\)	\(513\)

Input:

int((a+a*sin(f*x+e))^(5/2)/(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)*(((-12*cos(1/2*f*x+1/2*e)^4-16*cos(1/2*f* 
x+1/2*e)^2+43)*sin(1/2*f*x+1/2*e)+12*cos(1/2*f*x+1/2*e)^5-40*cos(1/2*f*x+1 
/2*e)^3-15*cos(1/2*f*x+1/2*e))*c^3+((24*cos(1/2*f*x+1/2*e)^6-136*cos(1/2*f 
*x+1/2*e)^4-158*cos(1/2*f*x+1/2*e)^2+14)*sin(1/2*f*x+1/2*e)+24*cos(1/2*f*x 
+1/2*e)^7+64*cos(1/2*f*x+1/2*e)^5-358*cos(1/2*f*x+1/2*e)^3+256*cos(1/2*f*x 
+1/2*e))*c^2*d+((112*cos(1/2*f*x+1/2*e)^6-652*cos(1/2*f*x+1/2*e)^4+484*cos 
(1/2*f*x+1/2*e)^2+3)*sin(1/2*f*x+1/2*e)+112*cos(1/2*f*x+1/2*e)^7+316*cos(1 
/2*f*x+1/2*e)^5-484*cos(1/2*f*x+1/2*e)^3+53*cos(1/2*f*x+1/2*e))*c*d^2+cos( 
1/2*f*x+1/2*e)*((-344*cos(1/2*f*x+1/2*e)^4-344*sin(1/2*f*x+1/2*e)*cos(1/2* 
f*x+1/2*e)^3+288*cos(1/2*f*x+1/2*e)^2+56*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^2/(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. 654 vs. \(2 (171) = 342\).

Time = 0.12 (sec) , antiderivative size = 654, normalized size of antiderivative = 3.46 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {2 \, {\left (32 \, a^{2} c^{2} - 64 \, a^{2} c d + 32 \, a^{2} d^{2} - {\left (3 \, a^{2} c^{2} + 14 \, a^{2} c d + 43 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (11 \, a^{2} c^{2} + 78 \, a^{2} c d - 29 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (23 \, a^{2} c^{2} + 14 \, a^{2} c d + 23 \, a^{2} d^{2}\right )} \cos \left (f x + e\right ) - {\left (32 \, a^{2} c^{2} - 64 \, a^{2} c d + 32 \, a^{2} d^{2} - {\left (3 \, a^{2} c^{2} + 14 \, a^{2} c d + 43 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (7 \, a^{2} c^{2} + 46 \, a^{2} c d + 7 \, a^{2} 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))^(5/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="fric 
as")
 

Output:

-2/15*(32*a^2*c^2 - 64*a^2*c*d + 32*a^2*d^2 - (3*a^2*c^2 + 14*a^2*c*d + 43 
*a^2*d^2)*cos(f*x + e)^3 + (11*a^2*c^2 + 78*a^2*c*d - 29*a^2*d^2)*cos(f*x 
+ e)^2 + 2*(23*a^2*c^2 + 14*a^2*c*d + 23*a^2*d^2)*cos(f*x + e) - (32*a^2*c 
^2 - 64*a^2*c*d + 32*a^2*d^2 - (3*a^2*c^2 + 14*a^2*c*d + 43*a^2*d^2)*cos(f 
*x + e)^2 - 2*(7*a^2*c^2 + 46*a^2*c*d + 7*a^2*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. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(5/2)/(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. 464 vs. \(2 (171) = 342\).

Time = 0.23 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.46 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {2 \, {\left ({\left (43 \, c^{3} + 14 \, c^{2} d + 3 \, c d^{2}\right )} a^{\frac {5}{2}} - \frac {{\left (15 \, c^{3} - 256 \, c^{2} d - 53 \, c d^{2} - 6 \, d^{3}\right )} a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {{\left (113 \, c^{3} - 116 \, c^{2} d + 493 \, c d^{2} + 50 \, d^{3}\right )} a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, {\left (17 \, c^{3} - 82 \, c^{2} d + 65 \, c d^{2} - 60 \, d^{3}\right )} a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, {\left (17 \, c^{3} - 82 \, c^{2} d + 65 \, c d^{2} - 60 \, d^{3}\right )} a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {{\left (113 \, c^{3} - 116 \, c^{2} d + 493 \, c d^{2} + 50 \, d^{3}\right )} a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {{\left (15 \, c^{3} - 256 \, c^{2} d - 53 \, c d^{2} - 6 \, d^{3}\right )} a^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {{\left (43 \, c^{3} + 14 \, c^{2} d + 3 \, c d^{2}\right )} a^{\frac {5}{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 )}}{15 \, {\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3} + \frac {{\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}}\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))^(5/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="maxi 
ma")
 

Output:

-2/15*((43*c^3 + 14*c^2*d + 3*c*d^2)*a^(5/2) - (15*c^3 - 256*c^2*d - 53*c* 
d^2 - 6*d^3)*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + (113*c^3 - 116*c^2* 
d + 493*c*d^2 + 50*d^3)*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*(1 
7*c^3 - 82*c^2*d + 65*c*d^2 - 60*d^3)*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) 
 + 1)^3 + 5*(17*c^3 - 82*c^2*d + 65*c*d^2 - 60*d^3)*a^(5/2)*sin(f*x + e)^4 
/(cos(f*x + e) + 1)^4 - (113*c^3 - 116*c^2*d + 493*c*d^2 + 50*d^3)*a^(5/2) 
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + (15*c^3 - 256*c^2*d - 53*c*d^2 - 6* 
d^3)*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - (43*c^3 + 14*c^2*d + 3* 
c*d^2)*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*(sin(f*x + e)^2/(cos(f 
*x + e) + 1)^2 + 1)/((c^3 + 3*c^2*d + 3*c*d^2 + d^3 + (c^3 + 3*c^2*d + 3*c 
*d^2 + d^3)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(c + 2*d*sin(f*x + e)/(co 
s(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)^(7/2)*f)
                                                                                    
                                                                                    
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="giac 
")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 27.70 (sec) , antiderivative size = 590, normalized size of antiderivative = 3.12 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx =\text {Too large to display} \] Input:

int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x))^(7/2),x)
 

Output:

-((c + d*sin(e + f*x))^(1/2)*((8*a^2*exp(e*4i + f*x*4i)*(a + a*sin(e + f*x 
))^(1/2)*(15*c^2 - 10*c*d + 7*d^2))/(3*d^3*f*(c + d)^3) + (4*a^2*exp(e*2i 
+ f*x*2i)*(a + a*sin(e + f*x))^(1/2)*(34*c*d + 5*c^2 - 3*d^2))/(3*d^3*f*(c 
 + d)^3) - (8*a^2*exp(e*3i + f*x*3i)*(a + a*sin(e + f*x))^(1/2)*(c^2*15i - 
 c*d*10i + d^2*7i))/(3*d^3*f*(c + d)^3) - (4*a^2*exp(e*5i + f*x*5i)*(a + a 
*sin(e + f*x))^(1/2)*(c*d*34i + c^2*5i - d^2*3i))/(3*d^3*f*(c + d)^3) - (4 
*a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^(1/2)*(14*c*d + 3*c^2 + 43*d^ 
2))/(15*d^3*f*(c + d)^3) + (4*a^2*exp(e*1i + f*x*1i)*(a + a*sin(e + f*x))^ 
(1/2)*(c*d*14i + c^2*3i + d^2*43i))/(15*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]

\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx=\sqrt {a}\, a^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{4} d^{4}+4 \sin \left (f x +e \right )^{3} c \,d^{3}+6 \sin \left (f x +e \right )^{2} c^{2} d^{2}+4 \sin \left (f x +e \right ) c^{3} d +c^{4}}d x +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4} d^{4}+4 \sin \left (f x +e \right )^{3} c \,d^{3}+6 \sin \left (f x +e \right )^{2} c^{2} d^{2}+4 \sin \left (f x +e \right ) c^{3} d +c^{4}}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4} d^{4}+4 \sin \left (f x +e \right )^{3} c \,d^{3}+6 \sin \left (f x +e \right )^{2} c^{2} d^{2}+4 \sin \left (f x +e \right ) c^{3} d +c^{4}}d x \right ) \] Input:

int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(7/2),x)
 

Output:

sqrt(a)*a**2*(int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + 
 f*x)**2)/(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) + 2*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*s 
in(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))