Integrand size = 29, antiderivative size = 188 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {2 d^{3/2} \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{3 \sqrt {3} f}-\frac {\sqrt {c-d} (c+5 d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{6 \sqrt {6} f}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (3+3 \sin (e+f x))^{3/2}} \]
-2*d^(3/2)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*s in(f*x+e))^(1/2))/a^(3/2)/f-1/4*(c+5*d)*arctanh(1/2*cos(f*x+e)*a^(1/2)*(c- d)^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))*(c-d)^(1/2 )/a^(3/2)/f*2^(1/2)-1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin (f*x+e))^(3/2)
Result contains complex when optimal does not.
Time = 15.34 (sec) , antiderivative size = 1625, normalized size of antiderivative = 8.64 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx =\text {Too large to display} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*((-c + d)/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (c*Sin[(e + f*x)/2] - d*Sin[(e + f*x)/2])/(Cos[(e + f *x)/2] + Sin[(e + f*x)/2])^2)*Sqrt[c + d*Sin[e + f*x]])/(f*(3 + 3*Sin[e + f*x])^(3/2)) + ((Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*Log[1 + Tan[(e + f*x)/2]] - Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]] - (4*I) *Sqrt[c - d]*d^(3/2)*(Log[(c - I*(d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Co s[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x)/2 ])/(2*d^(5/2)*(I + Tan[(e + f*x)/2]))] - Log[(c + I*d + (1 + I)*Sqrt[2]*Sq rt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d)*T an[(e + f*x)/2])/(2*d^(5/2)*(-I + Tan[(e + f*x)/2]))]))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(c^2/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (c*d)/((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) - d^2/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (d^2*Sin[e + f*x])/((Cos[(e + f*x)/2] + Sin[(e + f*x)/ 2])*Sqrt[c + d*Sin[e + f*x]])))/(f*(3 + 3*Sin[e + f*x])^(3/2)*(((c^2 + 4*c *d - 5*d^2)*Sec[(e + f*x)/2]^2)/(Sqrt[2]*(1 + Tan[(e + f*x)/2])) - (Sqrt[2 ]*(c^2 + 4*c*d - 5*d^2)*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d]*d* Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + Sqr t[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e ...
Time = 0.94 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3244, 27, 3042, 3461, 3042, 3254, 218, 3261, 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.
\(\displaystyle \int \frac {(c+d \sin (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d \sin (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{3/2}}dx\) |
\(\Big \downarrow \) 3244 |
\(\displaystyle -\frac {\int -\frac {4 a \sin (e+f x) d^2+a \left (c^2+4 d c-d^2\right )}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 a \sin (e+f x) d^2+a \left (c^2+4 d c-d^2\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {4 a \sin (e+f x) d^2+a \left (c^2+4 d c-d^2\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3461 |
\(\displaystyle \frac {4 d^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx+a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 d^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx+a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3254 |
\(\displaystyle \frac {a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {8 a d^2 \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {8 \sqrt {a} d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \frac {-\frac {2 a^2 (c-d) (c+5 d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {8 \sqrt {a} d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {8 \sqrt {a} d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {\sqrt {2} \sqrt {a} \sqrt {c-d} (c+5 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
((-8*Sqrt[a]*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin [e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (Sqrt[2]*Sqrt[a]*Sqrt[c - d]*(c + 5*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin [e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f)/(4*a^2) - ((c - d)*Cos[e + f*x]* Sqrt[c + d*Sin[e + f*x]])/(2*f*(a + a*Sin[e + f*x])^(3/2))
3.6.95.3.1 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_.)*(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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (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[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[(A*b - a*B)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) , x], x] + Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]]/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] && 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. \(8332\) vs. \(2(159)=318\).
Time = 1.16 (sec) , antiderivative size = 8333, normalized size of antiderivative = 44.32
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (159) = 318\).
Time = 0.72 (sec) , antiderivative size = 2883, normalized size of antiderivative = 15.34 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(sqrt(1/2)*((a*c + 5*a*d)*cos(f*x + e)^2 - 2*a*c - 10*a*d - (a*c + 5* a*d)*cos(f*x + e) - (2*a*c + 10*a*d + (a*c + 5*a*d)*cos(f*x + e))*sin(f*x + e))*sqrt((c - d)/a)*log((4*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin (f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) - (c - 3* d)*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + e) - 2*c - 2*d)*sin(f*x + e) - 2*c - 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin (f*x + e) - cos(f*x + e) - 2)) + (a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x + e) + 2*a*d)*sin(f*x + e))*sqrt(-d/a)*log((128*d^4*c os(f*x + e)^5 + 128*(2*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2 *d^2 + 4*c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 32*(c^3*d - 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 - 8*(16*d^3*cos(f *x + e)^4 + 24*(c*d^2 - d^3)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 31*c*d^2 - 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d + 59*c*d^2 - 51*d^3 - 8*(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 1 4*c*d^2 + 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqr t(d*sin(f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d ^3 + 289*d^4)*cos(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c ^2*d^2 + 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^2*d^2 - 6*c*d^3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - ...
\[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]