\(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\) [299]

Optimal result

Integrand size = 37, antiderivative size = 221 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=-\frac {a^{3/2} \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{5/2} (c+d)^{5/2} f}+\frac {a (B c-A d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac {a^2 \left (A (c-5 d) d+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-1/4*a^(3/2)*(A*d*(c+7*d)+3*B*(c^2+3*c*d+4*d^2))*arctanh(a^(1/2)*d^(1/2)*c 
os(f*x+e)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d^(5/2)/(c+d)^(5/2)/f+1/2*a* 
(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d/(c+d)/f/(c+d*sin(f*x+e))^2+ 
1/4*a^2*(A*(c-5*d)*d+B*(3*c^2+5*c*d-4*d^2))*cos(f*x+e)/d^2/(c+d)^2/f/(a+a* 
sin(f*x+e))^(1/2)/(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 = 16.34 (sec) , antiderivative size = 957, normalized size of antiderivative = 4.33 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(3/2)*(((A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*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)^(7/2) + ((A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*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)^(7/2) - (8*Sqrt[d]*(-c + d)*(-(B*c) + A*d)*(Cos[(e + f*x 
)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x])^2) - (4*Sqrt[d]...
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {3042, 3454, 27, 3042, 3459, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\int -\frac {\sqrt {\sin (e+f x) a+a} (a (B (c-4 d)-5 A d)-a (3 B c+A d+4 B d) \sin (e+f x))}{2 (c+d \sin (e+f x))^2}dx}{2 d (c+d)}+\frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\int \frac {\sqrt {\sin (e+f x) a+a} (a (B (c-4 d)-5 A d)-a (3 B c+A d+4 B d) \sin (e+f x))}{(c+d \sin (e+f x))^2}dx}{4 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\int \frac {\sqrt {\sin (e+f x) a+a} (a (B (c-4 d)-5 A d)-a (3 B c+A d+4 B d) \sin (e+f x))}{(c+d \sin (e+f x))^2}dx}{4 d (c+d)}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {-\frac {a \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 d (c+d)}-\frac {a^2 \left (A d (c-5 d)+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {-\frac {a \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 d (c+d)}-\frac {a^2 \left (A d (c-5 d)+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 d (c+d)}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f (c+d)}-\frac {a^2 \left (A d (c-5 d)+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 d (c+d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^{3/2} \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}}-\frac {a^2 \left (A d (c-5 d)+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 d (c+d)}\)

Input:

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

Output:

(a*(B*c - A*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2*d*(c + d)*f*(c + 
d*Sin[e + f*x])^2) - ((a^(3/2)*(A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*d^2)) 
*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f* 
x]])])/(d^(3/2)*(c + d)^(3/2)*f) - (a^2*(A*(c - 5*d)*d + B*(3*c^2 + 5*c*d 
- 4*d^2))*Cos[e + f*x])/(d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e 
 + f*x])))/(4*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 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. \(894\) vs. \(2(197)=394\).

Time = 0.55 (sec) , antiderivative size = 895, normalized size of antiderivative = 4.05

method	result	size
default	\(\frac {\left (\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} d^{2} \left (A c d +7 A \,d^{2}+3 B \,c^{2}+9 B c d +12 B \,d^{2}\right ) \cos \left (f x +e \right )^{2}-2 \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c d \left (A c d +7 A \,d^{2}+3 B \,c^{2}+9 B c d +12 B \,d^{2}\right )+A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, c \,d^{2}+7 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, d^{3}-A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c^{3} d -7 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c^{2} d^{2}-A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c \,d^{3}-7 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} d^{4}-5 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, c^{2} d -7 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, c \,d^{2}+4 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, d^{3}-3 a^{2} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) B \,c^{4}-9 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c^{3} d -15 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c^{2} d^{2}-9 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c \,d^{3}-12 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} d^{4}+A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,c^{2} d -8 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a c \,d^{2}-9 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,d^{3}+3 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,c^{3}+12 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,c^{2} d +5 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a c \,d^{2}-4 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,d^{3}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (1+\sin \left (f x +e \right )\right )}{4 \sqrt {\left (c +d \right ) a d}\, \left (c +d \sin \left (f x +e \right )\right )^{2} \left (c +d \right )^{2} d^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(895\)

Input:

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x,method=_R 
ETURNVERBOSE)
 

Output:

1/4*(arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(A*c*d+ 
7*A*d^2+3*B*c^2+9*B*c*d+12*B*d^2)*cos(f*x+e)^2-2*sin(f*x+e)*arctanh((a-a*s 
in(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d*(A*c*d+7*A*d^2+3*B*c^2+9*B 
*c*d+12*B*d^2)+A*(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*c*d^2+7*A*(a-a*s 
in(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*d^3-A*arctanh((a-a*sin(f*x+e))^(1/2)*d/ 
(a*c*d+a*d^2)^(1/2))*a^2*c^3*d-7*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-A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2 
)^(1/2))*a^2*c*d^3-7*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2 
))*a^2*d^4-5*B*(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*c^2*d-7*B*(a-a*sin 
(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*c*d^2+4*B*(a-a*sin(f*x+e))^(3/2)*((c+d)*a 
*d)^(1/2)*d^3-3*a^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))* 
B*c^4-9*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^3*d- 
15*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2-9*B 
*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^3-12*B*arct 
anh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^4+A*(a-a*sin(f*x+e 
))^(1/2)*((c+d)*a*d)^(1/2)*a*c^2*d-8*A*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^ 
(1/2)*a*c*d^2-9*A*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*d^3+3*B*(a-a* 
sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c^3+12*B*(a-a*sin(f*x+e))^(1/2)*((c+ 
d)*a*d)^(1/2)*a*c^2*d+5*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c*d^2 
-4*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*d^3)*(-a*(sin(f*x+e)-1)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 946 vs. \(2 (197) = 394\).

Time = 1.33 (sec) , antiderivative size = 2208, normalized size of antiderivative = 9.99 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, al 
gorithm="fricas")
 

Output:

[-1/16*((3*B*a*c^4 + (A + 15*B)*a*c^3*d + 3*(3*A + 11*B)*a*c^2*d^2 + 3*(5* 
A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4 - (3*B*a*c^2*d^2 + (A + 9*B)*a*c*d^ 
3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^3 - (6*B*a*c^3*d + (2*A + 21*B)*a*c^2 
*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^2 + (3*B* 
a*c^4 + (A + 9*B)*a*c^3*d + (7*A + 15*B)*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + ( 
7*A + 12*B)*a*d^4)*cos(f*x + e) + (3*B*a*c^4 + (A + 15*B)*a*c^3*d + 3*(3*A 
 + 11*B)*a*c^2*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4 - (3*B*a* 
c^2*d^2 + (A + 9*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^2 + 2*(3*B* 
a*c^3*d + (A + 9*B)*a*c^2*d^2 + (7*A + 12*B)*a*c*d^3)*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))*sqrt(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*cos(f*x 
+ e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(3*B*a 
*c^3 + (A + 2*B)*a*c^2*d - 3*(2*A + 3*B)*a*c*d^2 + (5*A + 4*B)*a*d^3 + (5* 
B*a*c^2*d - (A - 7*B)*a*c*d^2 - (7*A + 4*B)*a*d^3)*cos(f*x + e)^2 + (3*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, al 
gorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (197) = 394\).

Time = 0.29 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.82 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, al 
gorithm="giac")
 

Output:

-1/8*sqrt(2)*sqrt(a)*(sqrt(2)*(3*B*a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e 
)) + A*a*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 9*B*a*c*d*sgn(cos(-1/4* 
pi + 1/2*f*x + 1/2*e)) + 7*A*a*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 1 
2*B*a*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*p 
i + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((c^2*d^2 + 2*c*d^3 + d^4)*sqrt(-c* 
d - d^2)) - 2*(10*B*a*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*p 
i + 1/2*f*x + 1/2*e)^3 - 2*A*a*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*s 
in(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 14*B*a*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)^3 - 14*A*a*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 - 8*B*a*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 - 3*B*a*c^3*sgn(c 
os(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - A*a*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) - 12*B* 
a*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) 
 + 8*A*a*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) - 5*B*a*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) + 9*A*a*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) + 4*B*a*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^2*d^2 + 2*c*d^3 + d^4)*(2*d*sin(-1/4*pi + 1/ 
2*f*x + 1/2*e)^2 - c - d)^2))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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