Integrand size = 29, antiderivative size = 475 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\frac {2 (c-d) \sqrt {c+d} \left (12 b c-27 d-b^2 d\right ) E\left (\arcsin \left (\frac {\sqrt {3+b} \sqrt {c+d \sin (e+f x)}}{\sqrt {c+d} \sqrt {3+b \sin (e+f x)}}\right )|\frac {(3-b) (c+d)}{(3+b) (c-d)}\right ) \sec (e+f x) \sqrt {-\frac {(b c-3 d) (1-\sin (e+f x))}{(c+d) (3+b \sin (e+f x))}} \sqrt {\frac {(b c-3 d) (1+\sin (e+f x))}{(c-d) (3+b \sin (e+f x))}} (3+b \sin (e+f x))}{3 (3-b)^2 (3+b)^{3/2} (b c-3 d)^2 f}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 \left (9-b^2\right ) f (3+b \sin (e+f x))^{3/2}}+\frac {2 (9+b) (c-d) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {3+b \sin (e+f x)}}{\sqrt {3+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(3+b) (c-d)}{(3-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-3 d) (1-\sin (e+f x))}{(3+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-3 d) (1+\sin (e+f x))}{(3-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (3-b)^2 \sqrt {3+b} (b c-3 d) \sqrt {c+d} f} \]
2/3*(c-d)*(-3*a^2*d+4*a*b*c-b^2*d)*EllipticE((a+b)^(1/2)*(c+d*sin(f*x+e))^ (1/2)/(c+d)^(1/2)/(a+b*sin(f*x+e))^(1/2),((a-b)*(c+d)/(a+b)/(c-d))^(1/2))* sec(f*x+e)*(a+b*sin(f*x+e))*(c+d)^(1/2)*(-(-a*d+b*c)*(1-sin(f*x+e))/(c+d)/ (a+b*sin(f*x+e)))^(1/2)*((-a*d+b*c)*(1+sin(f*x+e))/(c-d)/(a+b*sin(f*x+e))) ^(1/2)/(a-b)^2/(a+b)^(3/2)/(-a*d+b*c)^2/f+2/3*(3*a+b)*(c-d)*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)*(c+d*sin(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)/(a-b)^2/(-a*d+b*c)/f/(a+b)^(1/2)/(c+d)^(1/2)+2/3*b*cos (f*x+e)*(c+d*sin(f*x+e))^(1/2)/(a^2-b^2)/f/(a+b*sin(f*x+e))^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(2011\) vs. \(2(475)=950\).
Time = 6.44 (sec) , antiderivative size = 2011, normalized size of antiderivative = 4.23 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \]
(Sqrt[3 + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]*((-2*b*Cos[e + f*x])/(3 *(-9 + b^2)*(3 + b*Sin[e + f*x])^2) - (2*(-12*b^2*c*Cos[e + f*x] + 27*b*d* Cos[e + f*x] + b^3*d*Cos[e + f*x]))/(3*(-9 + b^2)^2*(b*c - 3*d)*(3 + b*Sin [e + f*x]))))/f + ((-4*(-(b*c) + 3*d)*(27*b*c^2 + b^3*c^2 - 81*c*d - 3*b^2 *c*d + 9*b*d^2 - b^3*d^2)*Sqrt[((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcSin[Sqrt[((-3 - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[ e + f*x]))/(-(b*c) + 3*d)]/Sqrt[2]], (2*(-(b*c) + 3*d))/((3 + 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*(3 + b*Sin[e + f*x]))/(-(b*c) + 3*d)]*Sqrt[((-3 - b)*Csc[(-e + Pi /2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + 3*d)])/((3 + b)*(c + d)*Sqr t[3 + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - 4*(-(b*c) + 3*d)*(12*b^2 *c^2 + 9*b*c*d - b^3*c*d - 81*d^2 - 3*b^2*d^2)*((Sqrt[((c + d)*Cot[(-e + P i/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcSin[Sqrt[((-3 - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + 3*d)]/Sqrt[2]], (2*(-(b*c) + 3 *d))/((3 + 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*(3 + b*Sin[e + f*x]))/(-(b*c) + 3*d)]*Sqrt [((-3 - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + 3*d) ])/((3 + b)*(c + d)*Sqrt[3 + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - ( Sqrt[((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticPi[(-(b*c) + 3 *d)/((3 + b)*d), ArcSin[Sqrt[((-3 - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + ...
Time = 1.35 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3275, 27, 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.
| \(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{(a+b \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{(a+b \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3275 |
| \(\displaystyle \frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}-\frac {2 \int -\frac {3 a c-b d-(b c-3 a d) \sin (e+f x)}{2 (a+b \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {3 a c-b d-(b c-3 a d) \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{3 \left (a^2-b^2\right )}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a c-b d-(b c-3 a d) \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{3 \left (a^2-b^2\right )}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \frac {\frac {(3 a+b) (c-d) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{a-b}-\frac {\left (-3 a^2 d+4 a b c-b^2 d\right ) \int \frac {\sin (e+f x)+1}{(a+b \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{a-b}}{3 \left (a^2-b^2\right )}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(3 a+b) (c-d) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{a-b}-\frac {\left (-3 a^2 d+4 a b c-b^2 d\right ) \int \frac {\sin (e+f x)+1}{(a+b \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{a-b}}{3 \left (a^2-b^2\right )}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3297 |
| \(\displaystyle \frac {\frac {2 \sqrt {a+b} (3 a+b) (c-d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (a-b) \sqrt {c+d} (b c-a d)}-\frac {\left (-3 a^2 d+4 a b c-b^2 d\right ) \int \frac {\sin (e+f x)+1}{(a+b \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{a-b}}{3 \left (a^2-b^2\right )}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3475 |
| \(\displaystyle \frac {\frac {2 (c-d) \sqrt {c+d} \left (-3 a^2 d+4 a b c-b^2 d\right ) \sec (e+f x) (a+b \sin (e+f x)) \sqrt {-\frac {(b c-a d) (1-\sin (e+f x))}{(c+d) (a+b \sin (e+f x))}} \sqrt {\frac {(b c-a d) (\sin (e+f x)+1)}{(c-d) (a+b \sin (e+f x))}} E\left (\arcsin \left (\frac {\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}{\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}\right )|\frac {(a-b) (c+d)}{(a+b) (c-d)}\right )}{f (a-b) \sqrt {a+b} (b c-a d)^2}+\frac {2 \sqrt {a+b} (3 a+b) (c-d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (a-b) \sqrt {c+d} (b c-a d)}}{3 \left (a^2-b^2\right )}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}\) |
(2*b*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^(3/2)) + ((2*(c - d)*Sqrt[c + d]*(4*a*b*c - 3*a^2*d - b^2*d)*Ellip ticE[ArcSin[(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]])/(Sqrt[c + d]*Sqrt[a + b *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]))/((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]))]*(a + b*Sin[ e + f*x]))/((a - b)*Sqrt[a + b]*(b*c - a*d)^2*f) + (2*Sqrt[a + b]*(3*a + b )*(c - d)*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*Sin[e + f*x]) ))]*(c + d*Sin[e + f*x]))/((a - b)*Sqrt[c + d]*(b*c - a*d)*f))/(3*(a^2 - b ^2))
3.8.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^(m + 1)*((c + d*Sin[e + f*x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^ (n - 1)*Simp[a*c*(m + 1) + b*d*n + (a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(m + n + 2)*Sin[e + f*x]^2, 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] && LtQ[m, -1] && LtQ[0, n, 1] && IntegersQ[2*m, 2*n]
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)]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(175819\) vs. \(2(449)=898\).
Time = 11.56 (sec) , antiderivative size = 175820, normalized size of antiderivative = 370.15
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(-sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/(3*a*b^2*cos(f *x + e)^2 - a^3 - 3*a*b^2 + (b^3*cos(f*x + e)^2 - 3*a^2*b - b^3)*sin(f*x + e)), x)
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {c + d \sin {\left (e + f x \right )}}}{\left (a + b \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+b \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]