Integrand size = 29, antiderivative size = 405 \[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\frac {2 (a-b) \sqrt {a+b} d E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{(c-d) \sqrt {c+d} (b c-a d)^2 f}+\frac {2 \sqrt {a+b} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{(c-d) \sqrt {c+d} (b c-a d) f} \] Output:
2*(a-b)*(a+b)^(1/2)*d*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)/(c+d)^(1/2)/ (-a*d+b*c)^2/f+2*(a+b)^(1/2)*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)/(c+d) ^(1/2)/(-a*d+b*c)/f
Leaf count is larger than twice the leaf count of optimal. \(88729\) vs. \(2(405)=810\).
Time = 36.11 (sec) , antiderivative size = 88729, normalized size of antiderivative = 219.08 \[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[1/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]
Output:
Result too large to show
Time = 0.93 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3280, 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.
\(\displaystyle \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3280 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{c-d}-\frac {d \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{c-d}-\frac {d \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}\) |
\(\Big \downarrow \) 3297 |
\(\displaystyle \frac {2 \sqrt {a+b} \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}-\frac {d \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}\) |
\(\Big \downarrow \) 3475 |
\(\displaystyle \frac {2 \sqrt {a+b} \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}+\frac {2 d (a-b) \sqrt {a+b} \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)^2}\) |
Input:
Int[1/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]
Output:
(2*(a - b)*Sqrt[a + b]*d*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))]*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)^ 2*f) + (2*Sqrt[a + b]*EllipticF[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))]*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*S in[e + f*x])))]*(c + d*Sin[e + f*x]))/((c - d)*Sqrt[c + d]*(b*c - a*d)*f)
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin [(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[1/(a - b) Int[1/(Sqrt[a + b*Sin [e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Simp[b/(a - b) Int[(1 + Si n[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] / ; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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)]
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)]
Leaf count of result is larger than twice the leaf count of optimal. \(37683\) vs. \(2(375)=750\).
Time = 11.97 (sec) , antiderivative size = 37684, normalized size of antiderivative = 93.05
Input:
int(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x,method=_RETURNVERBOS E)
Output:
result too large to display
\[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="fr icas")
Output:
integral(sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/(a*c^2 + 2*b*c* d + a*d^2 - (2*b*c*d + a*d^2)*cos(f*x + e)^2 - (b*d^2*cos(f*x + e)^2 - b*c ^2 - 2*a*c*d - b*d^2)*sin(f*x + e)), x)
\[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {a + b \sin {\left (e + f x \right )}} \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(a+b*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**(3/2),x)
Output:
Integral(1/(sqrt(a + b*sin(e + f*x))*(c + d*sin(e + f*x))**(3/2)), x)
\[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="ma xima")
Output:
integrate(1/(sqrt(b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2)), x)
\[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="gi ac")
Output:
integrate(1/(sqrt(b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2)), x)
Timed out. \[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {a+b\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int(1/((a + b*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^(3/2)),x)
Output:
int(1/((a + b*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^(3/2)), x)
\[ \int \frac {1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right ) b +a}}{\sin \left (f x +e \right )^{3} b \,d^{2}+\sin \left (f x +e \right )^{2} a \,d^{2}+2 \sin \left (f x +e \right )^{2} b c d +2 \sin \left (f x +e \right ) a c d +\sin \left (f x +e \right ) b \,c^{2}+a \,c^{2}}d x \] Input:
int(1/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x)
Output:
int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x)*b + a))/(sin(e + f*x)**3*b *d**2 + sin(e + f*x)**2*a*d**2 + 2*sin(e + f*x)**2*b*c*d + 2*sin(e + f*x)* a*c*d + sin(e + f*x)*b*c**2 + a*c**2),x)