Integrand size = 29, antiderivative size = 172 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=-\frac {9 \sqrt {3} \left (3 c^2-10 c d+19 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{4 d^{5/2} f}+\frac {81 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 d^2 f \sqrt {3+3 \sin (e+f x)}}-\frac {9 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}{2 d f} \]
-1/4*a^(5/2)*(3*c^2-10*c*d+19*d^2)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a* sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/d^(5/2)/f+3/4*a^3*(c-3*d)*cos(f* x+e)*(c+d*sin(f*x+e))^(1/2)/d^2/f/(a+a*sin(f*x+e))^(1/2)-1/2*a^2*cos(f*x+e )*(a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(1/2)/d/f
Time = 2.99 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.51 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\frac {9 \sqrt {3} (1+\sin (e+f x))^{5/2} \left (\frac {\left (3 c^2-10 c d+19 d^2\right ) \left (2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )\right )}{d^{5/2}}+\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (3 c-11 d-2 d \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{d^2}\right )}{8 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
(9*Sqrt[3]*(1 + Sin[e + f*x])^(5/2)*(((3*c^2 - 10*c*d + 19*d^2)*(2*ArcTan[ (Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + Ar cTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]] ] - Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x ]]]))/d^(5/2) + (2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*c - 11*d - 2*d *Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]])/d^2))/(8*f*(Cos[(e + f*x)/2] + Si n[(e + f*x)/2])^5)
Time = 0.78 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3254, 218}
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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{\sqrt {c+d \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3242 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 (c+5 d)-3 a^2 (c-3 d) \sin (e+f x)\right )}{2 \sqrt {c+d \sin (e+f x)}}dx}{2 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 (c+5 d)-3 a^2 (c-3 d) \sin (e+f x)\right )}{\sqrt {c+d \sin (e+f x)}}dx}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 (c+5 d)-3 a^2 (c-3 d) \sin (e+f x)\right )}{\sqrt {c+d \sin (e+f x)}}dx}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {\frac {a^2 \left (3 c^2-10 c d+19 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{2 d}+\frac {3 a^3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a \sin (e+f x)+a}}}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a^2 \left (3 c^2-10 c d+19 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{2 d}+\frac {3 a^3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a \sin (e+f x)+a}}}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
\(\Big \downarrow \) 3254 |
\(\displaystyle \frac {\frac {3 a^3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a \sin (e+f x)+a}}-\frac {a^3 \left (3 c^2-10 c d+19 d^2\right ) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{d f}}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3 a^3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a \sin (e+f x)+a}}-\frac {a^{5/2} \left (3 c^2-10 c d+19 d^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}{2 d f}\) |
-1/2*(a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])/ (d*f) + (-((a^(5/2)*(3*c^2 - 10*c*d + 19*d^2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[ e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(3/2)*f )) + (3*a^3*(c - 3*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(d*f*Sqrt[a + a*Sin[e + f*x]]))/(4*d)
3.6.82.3.1 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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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 - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[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]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {c +d \sin \left (f x +e \right )}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (152) = 304\).
Time = 0.65 (sec) , antiderivative size = 1219, normalized size of antiderivative = 7.09 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \]
[1/32*((3*a^2*c^2 - 10*a^2*c*d + 19*a^2*d^2 + (3*a^2*c^2 - 10*a^2*c*d + 19 *a^2*d^2)*cos(f*x + e) + (3*a^2*c^2 - 10*a^2*c*d + 19*a^2*d^2)*sin(f*x + e ))*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2* d^2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a *c^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 32*(a*c^3*d - 2*a*c^2*d ^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4*cos(f*x + e)^4 - c^3* d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*( 5*c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3*d - 7*c^2*d^2 + 31*c* d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 17*c^2*d^2 + 59*c*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 1 4*c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqr t(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 4 76*a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f *x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*cos(f*x + e)^2 + 32*(a* c^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(c os(f*x + e) + sin(f*x + e) + 1)) - 8*(2*a^2*d*cos(f*x + e)^2 - 3*a^2*c + 9 *a^2*d - (3*a^2*c - 11*a^2*d)*cos(f*x + e) + (2*a^2*d*cos(f*x + e) + 3*a^2 *c - 9*a^2*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(d^2*f*cos(f*x + e) + d^2*f*sin(f*x + e) + d^2*f), 1/16*((3*a^2*c^...
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \]