Integrand size = 27, antiderivative size = 117 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=-\frac {3 \sqrt {3} (c+3 d) \text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {3+3 \sin (e+f x)}}\right )}{d^{3/2} (c+d)^{3/2} f}+\frac {9 (c-d) \cos (e+f x)}{d (c+d) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))} \]
-a^(3/2)*(c+3*d)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f *x+e))^(1/2))/d^(3/2)/(c+d)^(3/2)/f+a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*si n(f*x+e))/(a+a*sin(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 7.62 (sec) , antiderivative size = 762, normalized size of antiderivative = 6.51 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=\frac {3 \sqrt {3} (1+\sin (e+f x))^{3/2} \left ((c+3 d) \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-c \sqrt {d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-d^{3/2} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-d \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-2 c \sqrt {d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 d^{3/2} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-c \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+c \sqrt {d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+d^{3/2} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+3 d \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]+(c+3 d) \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-c \sqrt {d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-d^{3/2} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )+d \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-2 c \sqrt {d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 d^{3/2} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+c \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+c \sqrt {d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+d^{3/2} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-3 d \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+c \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]-\frac {4 \sqrt {d} \left (-c^2+d^2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{c+d \sin (e+f x)}\right )}{4 d^{3/2} (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]
(3*Sqrt[3]*(1 + Sin[e + f*x])^(3/2)*((c + 3*d)*RootSum[c + 4*d*#1 + 2*c*#1 ^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^( 3/2)*Log[-#1 + Tan[(e + f*x)/4]] - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4 ]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[ (e + f*x)/4]]*#1 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d ]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*# 1^2 + 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log [-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] + (c + 3*d)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Lo g[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] + d*Sqrt[ c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4 ]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/ 2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f *x)/4]]*#1^2 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] - (4*Sqrt[d]*(-c^2 + d^2)*(Cos[(e + f*x)/2] - Sin [(e + f*x)/2]))/(c + d*Sin[e + f*x])))/(4*d^(3/2)*(c + d)^2*f*(Cos[(e + f* x)/2] + Sin[(e + f*x)/2])^3)
Time = 0.49 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3241, 27, 2011, 3042, 3252, 221}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c+d \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c+d \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3241 |
| \(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {a \int -\frac {a (c+3 d)+a \sin (e+f x) (c+3 d)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{d (c+d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {a (c+3 d)+a \sin (e+f x) (c+3 d)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {a (c+3 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (c+3 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {a^2 (c+3 d) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f (c+d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {a^{3/2} (c+3 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}}\) |
-((a^(3/2)*(c + 3*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*S qrt[a + a*Sin[e + f*x]])])/(d^(3/2)*(c + d)^(3/2)*f)) + (a^2*(c - d)*Cos[e + f*x])/(d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))
3.6.34.3.1 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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[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_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*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]
Leaf count of result is larger than twice the leaf count of optimal. \(232\) vs. \(2(103)=206\).
Time = 1.08 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.99
| method | result | size |
| default | \(\frac {a \left (\sin \left (f x +e \right )+1\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-\sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a d \left (c +3 d \right )-\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c d +\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, c -\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, d \right )}{\left (c +d \right ) d \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(233\) |
a*(sin(f*x+e)+1)*(-a*(sin(f*x+e)-1))^(1/2)*(-sin(f*x+e)*arctanh((a-a*sin(f *x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*d*(c+3*d)-arctanh((a-a*sin(f*x+e))^( 1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^2-3*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c* d+a*d^2)^(1/2))*a*c*d+(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c-(a-a*sin( f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*d)/(c+d)/d/(c+d*sin(f*x+e))/(a*(c+d)*d)^(1 /2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (103) = 206\).
Time = 0.36 (sec) , antiderivative size = 970, normalized size of antiderivative = 8.29 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \]
[-1/4*((a*c^2 + 4*a*c*d + 3*a*d^2 - (a*c*d + 3*a*d^2)*cos(f*x + e)^2 + (a* c^2 + 3*a*c*d)*cos(f*x + e) + (a*c^2 + 4*a*c*d + 3*a*d^2 + (a*c*d + 3*a*d^ 2)*cos(f*x + e))*sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e) ^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2 *d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2 *d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e) )*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8* a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e )^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f *x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*si n(f*x + e))) + 4*(a*c - a*d + (a*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f *x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^2 + d^3)*f*cos(f*x + e)^2 - (c^2* d + c*d^2)*f*cos(f*x + e) - (c^2*d + 2*c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*c os(f*x + e) + (c^2*d + 2*c*d^2 + d^3)*f)*sin(f*x + e)), 1/2*((a*c^2 + 4*a* c*d + 3*a*d^2 - (a*c*d + 3*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 3*a*c*d)*cos(f *x + e) + (a*c^2 + 4*a*c*d + 3*a*d^2 + (a*c*d + 3*a*d^2)*cos(f*x + e))*sin (f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin (f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(a*c - a*d + (a*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f*x + e))*sqrt(a*sin(f*x ...
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}} \,d x } \]
Time = 0.37 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.68 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c d + d^{2}\right )} \sqrt {-c d - d^{2}}} - \frac {2 \, {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )} {\left (c d + d^{2}\right )}}\right )}}{2 \, f} \]
-1/2*sqrt(2)*sqrt(a)*(sqrt(2)*(a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3 *a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1 /2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((c*d + d^2)*sqrt(-c*d - d^2)) - 2*(a*c* sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - a*d*s gn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((2*d*s in(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)*(c*d + d^2)))/f
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]