Integrand size = 29, antiderivative size = 300 \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=-\frac {9 \sqrt {3} (c+d)^2 \left (3 c^2-26 c d+163 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{64 d^{5/2} f}-\frac {27 (c+d) \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d^2 f \sqrt {3+3 \sin (e+f x)}}-\frac {9 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{32 d^2 f \sqrt {3+3 \sin (e+f x)}}+\frac {9 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{8 d^2 f \sqrt {3+3 \sin (e+f x)}}-\frac {9 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f} \]
-1/64*a^(5/2)*(c+d)^2*(3*c^2-26*c*d+163*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-1/96*a^3*(3* c^2-26*c*d+163*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d^2/f/(a+a*sin(f*x+e ))^(1/2)+1/24*a^3*(3*c-17*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(5/2)/d^2/f/(a+a* sin(f*x+e))^(1/2)-1/4*a^2*cos(f*x+e)*(c+d*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e ))^(1/2)/d/f-1/64*a^3*(c+d)*(3*c^2-26*c*d+163*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 = 6.56 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.10 \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=\frac {9 \sqrt {3} (1+\sin (e+f x))^{5/2} \left (\frac {(c+d)^2 \left (3 c^2-26 c d+163 d^2\right ) \left (2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )\right )}{d^{5/2}}+\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)} \left (9 c^3-63 c^2 d-773 c d^2-581 d^3+4 d^2 (9 c+23 d) \cos (2 (e+f x))-2 d \left (3 c^2+158 c d+181 d^2\right ) \sin (e+f x)+12 d^3 \sin (3 (e+f x))\right )}{3 d^2}\right )}{128 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
(9*Sqrt[3]*(1 + Sin[e + f*x])^(5/2)*(((c + d)^2*(3*c^2 - 26*c*d + 163*d^2) *(2*ArcTan[(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*Si n[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*Sin[e + f*x]]*(9*c^3 - 63*c^2*d - 773*c*d^2 - 581*d^3 + 4*d^2*(9*c + 23*d)*Cos[2*(e + f*x)] - 2*d*(3*c^2 + 158*c*d + 181*d^2)*Sin[e + f*x] + 12 *d^3*Sin[3*(e + f*x)]))/(3*d^2)))/(128*f*(Cos[(e + f*x)/2] + Sin[(e + f*x) /2])^5)
Time = 1.22 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3249, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {\sin (e+f x) a+a} \left (a^2 (c+13 d)-a^2 (3 c-17 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}dx}{4 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\sin (e+f x) a+a} \left (a^2 (c+13 d)-a^2 (3 c-17 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}dx}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\sin (e+f x) a+a} \left (a^2 (c+13 d)-a^2 (3 c-17 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}dx}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}dx}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}dx}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \left (\frac {3}{4} (c+d) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \left (\frac {3}{4} (c+d) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \left (\frac {3}{4} (c+d) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \left (\frac {3}{4} (c+d) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \left (\frac {3}{4} (c+d) \left (-\frac {a (c+d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}+\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}+\frac {a^2 \left (3 c^2-26 c d+163 d^2\right ) \left (\frac {3}{4} (c+d) \left (-\frac {\sqrt {a} (c+d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {d} f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}}{8 d}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f}\) |
-1/4*(a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2) )/(d*f) + ((a^3*(3*c - 17*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(3*d *f*Sqrt[a + a*Sin[e + f*x]]) + (a^2*(3*c^2 - 26*c*d + 163*d^2)*(-1/2*(a*Co s[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(f*Sqrt[a + a*Sin[e + f*x]]) + (3*( c + d)*(-((Sqrt[a]*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[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))/(6*d))/(8*d )
3.6.80.3.1 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.
\[\int \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 646 vs. \(2 (274) = 548\).
Time = 1.03 (sec) , antiderivative size = 1751, normalized size of antiderivative = 5.84 \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=\text {Too large to display} \]
[1/1536*(3*(3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 1 63*a^2*d^4 + (3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 163*a^2*d^4)*cos(f*x + e) + (3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 163*a^2*d^4)*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)*co s(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))*s in(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*(48*a^2*d^3*cos(f*x + e)^4 + 9*a^2*c^3 - 57*a^2*c^2*d - 493*a^2*c*d^2 - 299*a^2*d^3 + 8*(9*a^2*c*d^2 + 23*a^2*d^3)*cos(f*x + e)^3 - 2*(3*a^2*...
Timed out. \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=\text {Timed out} \]
\[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]