\(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [305]

Optimal result

Integrand size = 37, antiderivative size = 265 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {a^{5/2} (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{7/2} (c+d)^{3/2} f}-\frac {a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{3 d^3 (c+d) f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 (5 B c-3 A d+2 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{d (c+d) f (c+d \sin (e+f x))} \] Output:

a^(5/2)*(c-d)*(A*d*(3*c+5*d)-B*(5*c^2+5*c*d-2*d^2))*arctanh(a^(1/2)*d^(1/2 
)*cos(f*x+e)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d^(7/2)/(c+d)^(3/2)/f-1/3 
*a^3*(3*A*d*(3*c+d)-B*(15*c^2-5*c*d-14*d^2))*cos(f*x+e)/d^3/(c+d)/f/(a+a*s 
in(f*x+e))^(1/2)-1/3*a^2*(-3*A*d+5*B*c+2*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^ 
(1/2)/d^2/(c+d)/f+a*(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/d/(c+d)/f 
/(c+d*sin(f*x+e))
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 15.29 (sec) , antiderivative size = 1002, normalized size of antiderivative = 3.78 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(5/2)*(-12*Sqrt[d]*(-4*B*c + 2*A*d + 5*B*d)*Cos[(e 
 + f*x)/2] - 4*B*d^(3/2)*Cos[(3*(e + f*x))/2] + (3*(c - d)*(-(A*d*(3*c + 5 
*d)) + B*(5*c^2 + 5*c*d - 2*d^2))*((c + d)*(e + f*x - 2*Log[Sec[(e + f*x)/ 
4]^2]) + Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , 
 (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f* 
x)/4]] - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + 
 Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 - c*Sqrt[ 
c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/ 
4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c + d]*Log[ 
-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*# 
1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c + d)^(5/2) + (3*(c - d)*(-(A 
*d*(3*c + 5*d)) + B*(5*c^2 + 5*c*d - 2*d^2))*(-((c + d)*(e + f*x - 2*Log[S 
ec[(e + f*x)/4]^2])) + Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^ 
3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 
 + Tan[(e + f*x)/4]] + d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqr 
t[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4] 
]*#1 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + 
Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - 3*d*Sq 
rt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + c*Sqrt[c + d]*Log[-#1 + Tan[( 
e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c + d)^(5/2) ...
 

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3454, 27, 3042, 3455, 27, 3042, 3460, 3042, 3252, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {2 \int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 \left (3 A (c-5 d) d-B \left (5 c^2-7 d c+6 d^2\right )\right )-a^2 \left (3 A d (3 c+d)-B \left (15 c^2-5 d c-14 d^2\right )\right ) \sin (e+f x)\right )}{2 (c+d \sin (e+f x))}dx}{3 d}+\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{2 d (c+d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 \left (3 A (c-5 d) d-B \left (5 c^2-7 d c+6 d^2\right )\right )-a^2 \left (3 A d (3 c+d)-B \left (15 c^2-5 d c-14 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)}dx}{3 d}+\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{2 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 \left (3 A (c-5 d) d-B \left (5 c^2-7 d c+6 d^2\right )\right )-a^2 \left (3 A d (3 c+d)-B \left (15 c^2-5 d c-14 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)}dx}{3 d}+\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{2 d (c+d)}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {3 a^2 (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}+\frac {2 a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}+\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{2 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {3 a^2 (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}+\frac {2 a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}+\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{2 d (c+d)}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {2 a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}-\frac {6 a^3 (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f}}{3 d}+\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{2 d (c+d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {2 a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}+\frac {\frac {2 a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}-\frac {6 a^{5/2} (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt {c+d}}}{3 d}}{2 d (c+d)}\)

Input:

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

Output:

(a*(B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(d*(c + d)*f*(c + 
d*Sin[e + f*x])) - ((2*a^2*(5*B*c - 3*A*d + 2*B*d)*Cos[e + f*x]*Sqrt[a + a 
*Sin[e + f*x]])/(3*d*f) + ((-6*a^(5/2)*(c - d)*(A*d*(3*c + 5*d) - B*(5*c^2 
 + 5*c*d - 2*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqr 
t[a + a*Sin[e + f*x]])])/(d^(3/2)*Sqrt[c + d]*f) + (2*a^3*(3*A*d*(3*c + d) 
 - B*(15*c^2 - 5*c*d - 14*d^2))*Cos[e + f*x])/(d*f*Sqrt[a + a*Sin[e + f*x] 
]))/(3*d))/(2*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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*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 3454

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

rule 3455

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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. \(931\) vs. \(2(241)=482\).

Time = 10.77 (sec) , antiderivative size = 932, normalized size of antiderivative = 3.52

method	result	size
default	\(\text {Expression too large to display}\)	\(932\)

Input:

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x,method=_R 
ETURNVERBOSE)
 

Output:

-1/3*a*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(sin(f*x+e)*d*(-9*A*arctan 
h((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d-6*A*arctanh((a-a 
*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^2+15*A*arctanh((a-a*sin( 
f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^3-2*B*(a-a*sin(f*x+e))^(3/2)*(( 
c+d)*a*d)^(1/2)*c*d-2*B*(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*d^2+15*a^ 
2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*B*c^3-21*B*arctanh 
((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^2+6*B*arctanh((a-a* 
sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^3+6*A*(a-a*sin(f*x+e))^(1/2 
)*((c+d)*a*d)^(1/2)*a*c*d+6*A*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*d 
^2-12*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c^2+6*B*(a-a*sin(f*x+e) 
)^(1/2)*((c+d)*a*d)^(1/2)*a*c*d+18*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1 
/2)*a*d^2)-9*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c 
^3*d-6*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2 
+15*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^3-2*B* 
(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*c^2*d-2*B*(a-a*sin(f*x+e))^(3/2)* 
((c+d)*a*d)^(1/2)*c*d^2+15*a^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d 
^2)^(1/2))*B*c^4-21*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2) 
)*a^2*c^2*d^2+6*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^ 
2*c*d^3+9*A*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c^2*d+3*A*(a-a*sin( 
f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*d^3-15*B*(a-a*sin(f*x+e))^(1/2)*((c+d...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 865 vs. \(2 (241) = 482\).

Time = 1.33 (sec) , antiderivative size = 2046, normalized size of antiderivative = 7.72 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, al 
gorithm="fricas")
 

Output:

[-1/12*(3*(5*B*a^2*c^4 - (3*A - 5*B)*a^2*c^3*d - (5*A + 7*B)*a^2*c^2*d^2 + 
 (3*A - 5*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4 - (5*B*a^2*c^3*d - 3*A*a^2*c^ 
2*d^2 - (2*A + 7*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4)*cos(f*x + e)^2 + (5*B 
*a^2*c^4 - 3*A*a^2*c^3*d - (2*A + 7*B)*a^2*c^2*d^2 + (5*A + 2*B)*a^2*c*d^3 
)*cos(f*x + e) + (5*B*a^2*c^4 - (3*A - 5*B)*a^2*c^3*d - (5*A + 7*B)*a^2*c^ 
2*d^2 + (3*A - 5*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4 + (5*B*a^2*c^3*d - 3*A 
*a^2*c^2*d^2 - (2*A + 7*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4)*cos(f*x + e))* 
sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a* 
c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3* 
d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e 
) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*s 
qrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)* 
cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c* 
d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^ 
2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*co 
s(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4* 
(15*B*a^2*c^3 - (9*A + 20*B)*a^2*c^2*d + 3*(2*A - 3*B)*a^2*c*d^2 + (3*A + 
14*B)*a^2*d^3 + 2*(B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x + e)^3 + 2*(5*B*a^2*c^ 
2*d - (3*A + 2*B)*a^2*c*d^2 - (3*A + 7*B)*a^2*d^3)*cos(f*x + e)^2 + (15*B* 
a^2*c^3 - (9*A + 10*B)*a^2*c^2*d - 15*B*a^2*c*d^2 - (3*A + 2*B)*a^2*d^3...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, al 
gorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (241) = 482\).

Time = 0.31 (sec) , antiderivative size = 597, normalized size of antiderivative = 2.25 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, al 
gorithm="giac")
 

Output:

-1/6*sqrt(2)*sqrt(a)*(3*sqrt(2)*(5*B*a^2*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1 
/2*e)) - 3*A*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*A*a^2*c*d^2 
*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 7*B*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2 
*f*x + 1/2*e)) + 5*A*a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*B*a^2 
*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1 
/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((c*d^3 + d^4)*sqrt(-c*d - d^2)) - 6*(B* 
a^2*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) 
 - A*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 
 1/2*e) - 2*B*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 
1/2*f*x + 1/2*e) + 2*A*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(- 
1/4*pi + 1/2*f*x + 1/2*e) + B*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
)*sin(-1/4*pi + 1/2*f*x + 1/2*e) - A*a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1 
/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((c*d^3 + d^4)*(2*d*sin(-1/4*pi + 1 
/2*f*x + 1/2*e)^2 - c - d)) + 4*(2*B*a^2*d^4*sgn(cos(-1/4*pi + 1/2*f*x + 1 
/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 6*B*a^2*c*d^3*sgn(cos(-1/4*pi + 
1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 3*A*a^2*d^4*sgn(cos(-1/ 
4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 9*B*a^2*d^4*sgn( 
cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/d^6)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**2*d**2 + 2*sin(e + 
 f*x)*c*d + c**2),x)*a + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin 
(e + f*x)**2*d**2 + 2*sin(e + f*x)*c*d + c**2),x)*b + int((sqrt(sin(e + f* 
x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**2*d**2 + 2*sin(e + f*x)*c*d + c**2 
),x)*a + 2*int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**2*d 
**2 + 2*sin(e + f*x)*c*d + c**2),x)*b + 2*int((sqrt(sin(e + f*x) + 1)*sin( 
e + f*x))/(sin(e + f*x)**2*d**2 + 2*sin(e + f*x)*c*d + c**2),x)*a + int((s 
qrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**2*d**2 + 2*sin(e + f*x) 
*c*d + c**2),x)*b)