Integrand size = 29, antiderivative size = 230 \[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=-\frac {9 \sqrt {3} (c+d) \left (c^2-6 c d+25 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{8 d^{5/2} f}-\frac {27 \left (c^2-6 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{8 d^2 f \sqrt {3+3 \sin (e+f x)}}+\frac {9 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 d^2 f \sqrt {3+3 \sin (e+f x)}}-\frac {3 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}{d f} \]
-1/8*a^(5/2)*(c+d)*(c^2-6*c*d+25*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+1/12*a^3*(3*c-13*d) *cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d^2/f/(a+a*sin(f*x+e))^(1/2)-1/3*a^2*co s(f*x+e)*(c+d*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/d/f-1/8*a^3*(c^2-6* c*d+25*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d^2/f/(a+a*sin(f*x+e))^(1/2)
Time = 2.38 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.25 \[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=\frac {9 \sqrt {3} (1+\sin (e+f x))^{5/2} \left (\frac {(c+d) \left (c^2-6 c d+25 d^2\right ) \left (2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )\right )}{d^{5/2}}+\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)} \left (3 c^2-16 c d-79 d^2+4 d^2 \cos (2 (e+f x))-2 d (c+17 d) \sin (e+f x)\right )}{3 d^2}\right )}{16 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
(9*Sqrt[3]*(1 + Sin[e + f*x])^(5/2)*(((c + d)*(c^2 - 6*c*d + 25*d^2)*(2*Ar cTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + ArcTanh[(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])*Sqrt[c + d*S in[e + f*x]]*(3*c^2 - 16*c*d - 79*d^2 + 4*d^2*Cos[2*(e + f*x)] - 2*d*(c + 17*d)*Sin[e + f*x]))/(3*d^2)))/(16*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) ^5)
Time = 0.98 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3249, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} \sqrt {c+d \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} \sqrt {c+d \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {\sin (e+f x) a+a} \left (a^2 (c+9 d)-a^2 (3 c-13 d) \sin (e+f x)\right ) \sqrt {c+d \sin (e+f x)}dx}{3 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\sin (e+f x) a+a} \left (a^2 (c+9 d)-a^2 (3 c-13 d) \sin (e+f x)\right ) \sqrt {c+d \sin (e+f x)}dx}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\sin (e+f x) a+a} \left (a^2 (c+9 d)-a^2 (3 c-13 d) \sin (e+f x)\right ) \sqrt {c+d \sin (e+f x)}dx}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {3 a^2 \left (c^2-6 c d+25 d^2\right ) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx}{4 d}+\frac {a^3 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt {a \sin (e+f x)+a}}}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 a^2 \left (c^2-6 c d+25 d^2\right ) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx}{4 d}+\frac {a^3 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt {a \sin (e+f x)+a}}}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {\frac {3 a^2 \left (c^2-6 c d+25 d^2\right ) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{4 d}+\frac {a^3 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt {a \sin (e+f x)+a}}}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 a^2 \left (c^2-6 c d+25 d^2\right ) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{4 d}+\frac {a^3 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt {a \sin (e+f x)+a}}}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {\frac {3 a^2 \left (c^2-6 c d+25 d^2\right ) \left (-\frac {a (c+d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{4 d}+\frac {a^3 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt {a \sin (e+f x)+a}}}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {a^3 (3 c-13 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt {a \sin (e+f x)+a}}+\frac {3 a^2 \left (c^2-6 c d+25 d^2\right ) \left (-\frac {\sqrt {a} (c+d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {d} f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{4 d}}{6 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}{3 d f}\) |
-1/3*(a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2) )/(d*f) + ((a^3*(3*c - 13*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(2*d *f*Sqrt[a + a*Sin[e + f*x]]) + (3*a^2*(c^2 - 6*c*d + 25*d^2)*(-((Sqrt[a]*( c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqr t[c + d*Sin[e + f*x]])])/(Sqrt[d]*f)) - (a*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])))/(4*d))/(6*d)
3.6.81.3.1 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_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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] && GtQ[n, 0] && IntegerQ[2*n]
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]
Timed out.
\[\int \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c +d \sin \left (f x +e \right )}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (209) = 418\).
Time = 0.76 (sec) , antiderivative size = 1455, normalized size of antiderivative = 6.33 \[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=\text {Too large to display} \]
[1/192*(3*(a^2*c^3 - 5*a^2*c^2*d + 19*a^2*c*d^2 + 25*a^2*d^3 + (a^2*c^3 - 5*a^2*c^2*d + 19*a^2*c*d^2 + 25*a^2*d^3)*cos(f*x + e) + (a^2*c^3 - 5*a^2*c ^2*d + 19*a^2*c*d^2 + 25*a^2*d^3)*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 + 5 1*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 - 14*c*d^3 + 13*d^4)*cos(f*x + e ))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a /d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 476*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))/(cos(f*x + e) + sin(f*x + e) + 1)) + 8*(8*a^2*d^2*cos(f*x + e)^3 + 3*a^2*c^2 - 14*a^2*c*d - 49*a^2*d^2 - 2*(a^2*c*d + 13*a^2*d^2)*cos(f*x + e)^2 + (3*a^2*c^2 - 16*a^2*c*d - 83*a^2 *d^2)*cos(f*x + e) - (8*a^2*d^2*cos(f*x + e)^2 + 3*a^2*c^2 - 14*a^2*c*d...
Timed out. \[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right ) + c} \,d x } \]
\[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right ) + c} \,d x } \]
Timed out. \[ \int (3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {c+d\,\sin \left (e+f\,x\right )} \,d x \]