\(\int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx\) [795]

Optimal result

Integrand size = 29, antiderivative size = 521 \[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 d^2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (a-b) \sqrt {a+b} d \left (2 a c d-b \left (3 c^2-d^2\right )\right ) E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (c-d)^2 (c+d)^{3/2} (b c-a d)^3 f}-\frac {2 \sqrt {a+b} \left (a d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (c-d)^2 (c+d)^{3/2} (b c-a d)^2 f} \] Output:

-2/3*d^2*cos(f*x+e)*(a+b*sin(f*x+e))^(1/2)/(-a*d+b*c)/(c^2-d^2)/f/(c+d*sin 
(f*x+e))^(3/2)-4/3*(a-b)*(a+b)^(1/2)*d*(2*a*c*d-b*(3*c^2-d^2))*EllipticE(( 
c+d)^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),((a+b 
)*(c-d)/(a-b)/(c+d))^(1/2))*sec(f*x+e)*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c 
+d*sin(f*x+e)))^(1/2)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a-b)/(c+d*sin(f*x+e)))^ 
(1/2)*(c+d*sin(f*x+e))/(c-d)^2/(c+d)^(3/2)/(-a*d+b*c)^3/f-2/3*(a+b)^(1/2)* 
(a*d*(3*c+d)-b*(3*c^2+3*c*d-2*d^2))*EllipticF((c+d)^(1/2)*(a+b*sin(f*x+e)) 
^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),((a+b)*(c-d)/(a-b)/(c+d))^(1/2)) 
*sec(f*x+e)*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^(1/2)*(-(-a 
*d+b*c)*(1+sin(f*x+e))/(a-b)/(c+d*sin(f*x+e)))^(1/2)*(c+d*sin(f*x+e))/(c-d 
)^2/(c+d)^(3/2)/(-a*d+b*c)^2/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2102\) vs. \(2(521)=1042\).

Time = 6.57 (sec) , antiderivative size = 2102, normalized size of antiderivative = 4.03 \[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[1/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)),x]
 

Output:

(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]*((-2*d^2*Cos[e + f*x])/ 
(3*(b*c - a*d)*(c^2 - d^2)*(c + d*Sin[e + f*x])^2) + (4*(-3*b*c^2*d^2*Cos[ 
e + f*x] + 2*a*c*d^3*Cos[e + f*x] + b*d^4*Cos[e + f*x]))/(3*(b*c - a*d)^2* 
(c^2 - d^2)^2*(c + d*Sin[e + f*x]))))/f + ((-4*(-(b*c) + a*d)*(3*b^2*c^4 - 
 6*a*b*c^3*d + 3*a^2*c^2*d^2 - 5*b^2*c^2*d^2 + 2*a*b*c*d^3 + a^2*d^4 + 2*b 
^2*d^4)*Sqrt[((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcS 
in[Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) 
 + a*d)]/Sqrt[2]], (2*(-(b*c) + a*d))/((a + b)*(-c + d))]*Sec[e + f*x]*Sin 
[(-e + Pi/2 - f*x)/2]^4*Sqrt[((c + d)*Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Si 
n[e + f*x]))/(-(b*c) + a*d)]*Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c 
+ d*Sin[e + f*x]))/(-(b*c) + a*d)])/((a + b)*(c + d)*Sqrt[a + b*Sin[e + f* 
x]]*Sqrt[c + d*Sin[e + f*x]]) - 4*(-(b*c) + a*d)*(-6*b^2*c^3*d - 2*a*b*c^2 
*d^2 + 4*a^2*c*d^3 + 2*b^2*c*d^3 + 2*a*b*d^4)*((Sqrt[((c + d)*Cot[(-e + Pi 
/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcSin[Sqrt[((-a - b)*Csc[(-e + Pi/2 - 
 f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + a*d)]/Sqrt[2]], (2*(-(b*c) + a* 
d))/((a + b)*(-c + d))]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[((c + 
 d)*Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/(-(b*c) + a*d)]*Sqrt[ 
((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + a*d)] 
)/((a + b)*(c + d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - (S 
qrt[((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticPi[(-(b*c) +...
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3281, 27, 25, 3042, 3477, 3042, 3297, 3475}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

\(\displaystyle \frac {\frac {2 d \left (2 a c d-b \left (3 c^2-d^2\right )\right ) \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}-\frac {\left (a d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{c-d}}{3 \left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 d \left (2 a c d-b \left (3 c^2-d^2\right )\right ) \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}-\frac {\left (a d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{c-d}}{3 \left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3297

\(\displaystyle \frac {\frac {2 d \left (2 a c d-b \left (3 c^2-d^2\right )\right ) \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}-\frac {2 \sqrt {a+b} \left (a d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}}{3 \left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3475

\(\displaystyle \frac {-\frac {2 \sqrt {a+b} \left (a d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}-\frac {4 d (a-b) \sqrt {a+b} \left (2 a c d-b \left (3 c^2-d^2\right )\right ) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)^2}}{3 \left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}\)

Input:

Int[1/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)),x]
 

Output:

(-2*d^2*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]])/(3*(b*c - a*d)*(c^2 - d^2)* 
f*(c + d*Sin[e + f*x])^(3/2)) + ((-4*(a - b)*Sqrt[a + b]*d*(2*a*c*d - b*(3 
*c^2 - d^2))*EllipticE[ArcSin[(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt 
[a + b]*Sqrt[c + d*Sin[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))]*S 
ec[e + f*x]*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d*Sin[e + 
f*x]))]*Sqrt[-(((b*c - a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f* 
x])))]*(c + d*Sin[e + f*x]))/((c - d)*Sqrt[c + d]*(b*c - a*d)^2*f) - (2*Sq 
rt[a + b]*(a*d*(3*c + d) - b*(3*c^2 + 3*c*d - 2*d^2))*EllipticF[ArcSin[(Sq 
rt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]]) 
], ((a + b)*(c - d))/((a - b)*(c + d))]*Sec[e + f*x]*Sqrt[((b*c - a*d)*(1 
- Sin[e + f*x]))/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 + 
Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*(c + d*Sin[e + f*x]))/((c 
- d)*Sqrt[c + d]*(b*c - a*d)*f))/(3*(b*c - a*d)*(c^2 - d^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

rule 3297

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ 
.) + (f_.)*(x_)]]), x_Symbol] :> Simp[2*((c + d*Sin[e + f*x])/(f*(b*c - a*d 
)*Rt[(c + d)/(a + b), 2]*Cos[e + f*x]))*Sqrt[(b*c - a*d)*((1 - Sin[e + f*x] 
)/((a + b)*(c + d*Sin[e + f*x])))]*Sqrt[(-(b*c - a*d))*((1 + Sin[e + f*x])/ 
((a - b)*(c + d*Sin[e + f*x])))]*EllipticF[ArcSin[Rt[(c + d)/(a + b), 2]*(S 
qrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]])], (a + b)*((c - d)/((a - 
b)*(c + d)))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && N 
eQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/(a + b)]
 

rule 3475

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_) + (b_.)*sin[(e_.) + (f_.) 
*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[-2*A*(c - d)*((a + b*Sin[e + f*x])/(f*(b*c - a*d)^2*Rt[(a + b)/(c + d), 2 
]*Cos[e + f*x]))*Sqrt[(b*c - a*d)*((1 + Sin[e + f*x])/((c - d)*(a + b*Sin[e 
 + f*x])))]*Sqrt[(-(b*c - a*d))*((1 - Sin[e + f*x])/((c + d)*(a + b*Sin[e + 
 f*x])))]*EllipticE[ArcSin[Rt[(a + b)/(c + d), 2]*(Sqrt[c + d*Sin[e + f*x]] 
/Sqrt[a + b*Sin[e + f*x]])], (a - b)*((c + d)/((a + b)*(c - d)))], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] 
&& NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(a + b)/(c + d)]
 

rule 3477

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(105867\) vs. \(2(481)=962\).

Time = 15.40 (sec) , antiderivative size = 105868, normalized size of antiderivative = 203.20

method	result	size
default	\(\text {Expression too large to display}\)	\(105868\)

Input:

int(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

result too large to display
 

Fricas [F]

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

integrate(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fr 
icas")
 

Output:

integral(sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/(b*d^3*cos(f*x 
+ e)^4 + a*c^3 + 3*b*c^2*d + 3*a*c*d^2 + b*d^3 - (3*b*c^2*d + 3*a*c*d^2 + 
2*b*d^3)*cos(f*x + e)^2 + (b*c^3 + 3*a*c^2*d + 3*b*c*d^2 + a*d^3 - (3*b*c* 
d^2 + a*d^3)*cos(f*x + e)^2)*sin(f*x + e)), x)
 

Sympy [F]

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

integrate(1/(a+b*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**(5/2),x)
 

Output:

Integral(1/(sqrt(a + b*sin(e + f*x))*(c + d*sin(e + f*x))**(5/2)), x)
 

Maxima [F]

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

integrate(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="ma 
xima")
 

Output:

integrate(1/(sqrt(b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2)), x)
 

Giac [F]

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

integrate(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="gi 
ac")
 

Output:

integrate(1/(sqrt(b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2)), x)
 

Mupad [F(-1)]

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

int(1/((a + b*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^(5/2)),x)
 

Output:

int(1/((a + b*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^(5/2)), x)
 

Reduce [F]

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

int(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x)
 

Output:

int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x)*b + a))/(sin(e + f*x)**4*b 
*d**3 + sin(e + f*x)**3*a*d**3 + 3*sin(e + f*x)**3*b*c*d**2 + 3*sin(e + f* 
x)**2*a*c*d**2 + 3*sin(e + f*x)**2*b*c**2*d + 3*sin(e + f*x)*a*c**2*d + si 
n(e + f*x)*b*c**3 + a*c**3),x)