\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx\) [600]

Optimal result

Integrand size = 29, antiderivative size = 131 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^{3/2} f}+\frac {2 d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(306\) vs. \(2(131)=262\).

Time = 3.65 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.34 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\frac {\frac {2 d \cos (e+f x)}{c+d}+\frac {\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}}{(c-d) f \sqrt {a (1+\sin (e+f x))} \sqrt {c+d \sin (e+f x)}} \] Input:

Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]
 

Output:

((2*d*Cos[e + f*x])/(c + d) + (Log[1 + Tan[(e + f*x)/2]] - Log[c - d + 2*S 
qrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + 
d)*Tan[(e + f*x)/2]])/(Sec[(e + f*x)/2]^2/(2 + 2*Tan[(e + f*x)/2]) - (-1/2 
*((c - d)*Sec[(e + f*x)/2]^2) + (Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/ 
2)*(d + d*Cos[e + f*x] + c*Sin[e + f*x]))/Sqrt[c + d*Sin[e + f*x]])/(c - d 
 + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + 
(-c + d)*Tan[(e + f*x)/2])))/((c - d)*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c 
+ d*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3258, 27, 3042, 3261, 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.

\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 d \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {\sqrt {2} (c+d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d} \left (c^2-d^2\right )}\)

Input:

Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]
 

Output:

-((Sqrt[2]*(c + d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqr 
t[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[a]*Sqrt[c - d]*(c^ 
2 - d^2)*f)) + (2*d*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]* 
Sqrt[c + d*Sin[e + f*x]])
 

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 3258

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

rule 3261

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(984\) vs. \(2(112)=224\).

Time = 10.14 (sec) , antiderivative size = 985, normalized size of antiderivative = 7.52

method	result	size
default	\(\text {Expression too large to display}\)	\(985\)

Input:

int(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

2/f/(2*c-2*d)^(1/2)/(c-d)/(c+d)*(((cos(1/2*f*x+1/2*e)+1)*sin(1/2*f*x+1/2*e 
)+cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)+1))*((c+2*d*sin(1/2*f*x+1/2*e)*co 
s(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*ln(-4*(((c+2*d*sin(1/2*f 
*x+1/2*e)*cos(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*(2*c-2*d)^(1 
/2)*cos(1/2*f*x+1/2*e)+(2*c-2*d)^(1/2)*((c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2* 
f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)-cos(1/2*f*x+1/2*e)*c+c*sin(1/2 
*f*x+1/2*e)+cos(1/2*f*x+1/2*e)*d-d*sin(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e) 
+sin(1/2*f*x+1/2*e)))*c^2+((2*cos(1/2*f*x+1/2*e)^2+1)*(cos(1/2*f*x+1/2*e)+ 
1)*sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e)*(2*cos(1/2*f*x+1/2*e)^2-3)*(cos(1 
/2*f*x+1/2*e)+1))*((c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))/(cos(1/2* 
f*x+1/2*e)+1)^2)^(1/2)*ln(-4*(((c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e 
))/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*(2*c-2*d)^(1/2)*cos(1/2*f*x+1/2*e)+(2*c 
-2*d)^(1/2)*((c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/ 
2*e)+1)^2)^(1/2)-cos(1/2*f*x+1/2*e)*c+c*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2 
*e)*d-d*sin(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+sin(1/2*f*x+1/2*e)))*c*d-2 
*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)+1)*(cos(1/2*f*x+1/2*e)^2-sin(1/2*f 
*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)*((c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2 
*e))/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*ln(-4*(((c+2*d*sin(1/2*f*x+1/2*e)*cos 
(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*(2*c-2*d)^(1/2)*cos(1/2*f 
*x+1/2*e)+(2*c-2*d)^(1/2)*((c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (112) = 224\).

Time = 0.41 (sec) , antiderivative size = 1016, normalized size of antiderivative = 7.76 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

[-1/4*(8*(d*cos(f*x + e) - d*sin(f*x + e) + d)*sqrt(a*sin(f*x + e) + a)*sq 
rt(d*sin(f*x + e) + c) - sqrt(2)*(a*c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2 
)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d^2 
 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*log(((c^2 - 14*c*d + 17*d^2 
)*cos(f*x + e)^3 - (13*c^2 - 22*c*d - 3*d^2)*cos(f*x + e)^2 + 4*sqrt(2)*(( 
c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^2 - 4*c^2 + 8*c*d - 4*d^2 - (3*c^2 - 4*c 
*d + d^2)*cos(f*x + e) + (4*c^2 - 8*c*d + 4*d^2 + (c^2 - 4*c*d + 3*d^2)*co 
s(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c 
)/sqrt(a*c - a*d) - 4*c^2 - 8*c*d - 4*d^2 - 2*(9*c^2 - 14*c*d + 9*d^2)*cos 
(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^ 
2 + 2*(7*c^2 - 18*c*d + 7*d^2)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 
 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 
 2*cos(f*x + e) - 4))/sqrt(a*c - a*d))/((a*c^2*d - a*d^3)*f*cos(f*x + e)^2 
 - (a*c^3 - a*c*d^2)*f*cos(f*x + e) - (a*c^3 + a*c^2*d - a*c*d^2 - a*d^3)* 
f - ((a*c^2*d - a*d^3)*f*cos(f*x + e) + (a*c^3 + a*c^2*d - a*c*d^2 - a*d^3 
)*f)*sin(f*x + e)), -1/2*(sqrt(2)*(a*c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^ 
2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d^ 
2 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-1/(a*c - a*d))*arcta 
n(-1/4*sqrt(2)*sqrt(a*sin(f*x + e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d) 
*sqrt(d*sin(f*x + e) + c)*sqrt(-1/(a*c - a*d))/(d*cos(f*x + e)*sin(f*x ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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