Integrand size = 29, antiderivative size = 513 \[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 d^2 \cos (e+f x) \sqrt {3+b \sin (e+f x)}}{3 (b c-3 d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (3-b) \sqrt {3+b} d \left (6 c d-b \left (3 c^2-d^2\right )\right ) E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {3+b \sin (e+f x)}}{\sqrt {3+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(3+b) (c-d)}{(3-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-3 d) (1-\sin (e+f x))}{(3+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-3 d) (1+\sin (e+f x))}{(3-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (b c-3 d)^3 (c-d)^2 (c+d)^{3/2} f}-\frac {2 \sqrt {3+b} \left (3 d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {3+b \sin (e+f x)}}{\sqrt {3+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(3+b) (c-d)}{(3-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-3 d) (1-\sin (e+f x))}{(3+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-3 d) (1+\sin (e+f x))}{(3-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (b c-3 d)^2 (c-d)^2 (c+d)^{3/2} f} \]
-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)*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)*(c+d*sin(f*x+e))*(a+b)^(1/2)*((-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)^2/(c+d)^(3/2)/(-a*d+b*c)^3/f-2/3*(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)* (c+d*sin(f*x+e))*(a+b)^(1/2)*((-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 )^2/(c+d)^(3/2)/(-a*d+b*c)^2/f
Leaf count is larger than twice the leaf count of optimal. \(2066\) vs. \(2(513)=1026\).
Time = 6.46 (sec) , antiderivative size = 2066, normalized size of antiderivative = 4.03 \[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \]
(Sqrt[3 + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]*((-2*d^2*Cos[e + f*x])/ (3*(b*c - 3*d)*(c^2 - d^2)*(c + d*Sin[e + f*x])^2) - (4*(3*b*c^2*d^2*Cos[e + f*x] - 6*c*d^3*Cos[e + f*x] - b*d^4*Cos[e + f*x]))/(3*(b*c - 3*d)^2*(c^ 2 - d^2)^2*(c + d*Sin[e + f*x]))))/f + ((-4*(-(b*c) + 3*d)*(3*b^2*c^4 - 18 *b*c^3*d + 27*c^2*d^2 - 5*b^2*c^2*d^2 + 6*b*c*d^3 + 9*d^4 + 2*b^2*d^4)*Sqr t[((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)]/Sq rt[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]]) - 4*(-(b*c) + 3*d)*(-6*b^2*c^3*d - 6*b*c^2*d^2 + 36*c*d ^3 + 2*b^2*c*d^3 + 6*b*d^4)*((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) *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),...
Time = 1.50 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.06, 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 |
\(\displaystyle \frac {-\frac {2 \sqrt {a+b} \left (a d (3 c+d)-b \left (3 c^2+3 c d-2 d^2\right )\right ) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}-\frac {4 d (a-b) \sqrt {a+b} \left (2 a c d-b \left (3 c^2-d^2\right )\right ) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)^2}}{3 \left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}\) |
(-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))
3.8.88.3.1 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^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])))
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. \(192752\) vs. \(2(481)=962\).
Time = 13.96 (sec) , antiderivative size = 192753, normalized size of antiderivative = 375.74
\[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
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)
\[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a + b \sin {\left (e + f x \right )}} \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {3+b \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a+b\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]