Integrand size = 25, antiderivative size = 141 \[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {a \arctan \left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {a \text {arctanh}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}} \]
2*(a*sin(f*x+e))^(1/2)/b/f/(b*tan(f*x+e))^(1/2)-a*arctan(cos(f*x+e)^(1/2)) *cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/b^2/f/(a*sin(f*x+e))^(1/2)-a*arctan h(cos(f*x+e)^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/b^2/f/(a*sin(f*x +e))^(1/2)
Time = 0.62 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {\left (-\arctan \left (\sqrt [4]{\cos ^2(e+f x)}\right )-\text {arctanh}\left (\sqrt [4]{\cos ^2(e+f x)}\right )+2 \sqrt [4]{\cos ^2(e+f x)}\right ) \sqrt {a \sin (e+f x)}}{b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \]
((-ArcTan[(Cos[e + f*x]^2)^(1/4)] - ArcTanh[(Cos[e + f*x]^2)^(1/4)] + 2*(C os[e + f*x]^2)^(1/4))*Sqrt[a*Sin[e + f*x]])/(b*f*(Cos[e + f*x]^2)^(1/4)*Sq rt[b*Tan[e + f*x]])
Time = 0.47 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.76, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3075, 3042, 3081, 27, 3042, 3045, 266, 756, 216, 219}
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.
| \(\displaystyle \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3075 |
| \(\displaystyle \frac {a^2 \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}}dx}{b^2}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}}dx}{b^2}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3081 |
| \(\displaystyle \frac {a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{a \sqrt {\cos (e+f x)}}dx}{b^2 \sqrt {a \sin (e+f x)}}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{\sqrt {\cos (e+f x)}}dx}{b^2 \sqrt {a \sin (e+f x)}}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \sin (e+f x)}dx}{b^2 \sqrt {a \sin (e+f x)}}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle \frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (1-\cos ^2(e+f x)\right )}d\cos (e+f x)}{b^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{1-\cos ^2(e+f x)}d\sqrt {\cos (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \int \frac {1}{\cos (e+f x)+1}d\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )\right )}{b^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (e+f x)}\right )\right )}{b^2 f \sqrt {a \sin (e+f x)}}\) |
(2*Sqrt[a*Sin[e + f*x]])/(b*f*Sqrt[b*Tan[e + f*x]]) - (2*a*(ArcTan[Sqrt[Co s[e + f*x]]]/2 + ArcTanh[Sqrt[Cos[e + f*x]]]/2)*Sqrt[Cos[e + f*x]]*Sqrt[b* Tan[e + f*x]])/(b^2*f*Sqrt[a*Sin[e + f*x]])
3.2.37.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]/((b_.)*tan[(e_.) + (f_.)*(x_)])^(3/ 2), x_Symbol] :> Simp[2*(Sqrt[a*Sin[e + f*x]]/(b*f*Sqrt[b*Tan[e + f*x]])), x] + Simp[a^2/b^2 Int[Sqrt[b*Tan[e + f*x]]/(a*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f}, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
Time = 1.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.47
| method | result | size |
| default | \(\frac {\left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )\right ) \sqrt {\sin \left (f x +e \right ) a}}{2 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \tan \left (f x +e \right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b}\) | \(207\) |
1/2/f*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+4*(-cos(f*x+e)/(c os(f*x+e)+1)^2)^(1/2)+arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-ln( (2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x +e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1)))*(sin(f*x+e)*a)^(1/2)/(cos(f *x+e)+1)/(b*tan(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/b
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (121) = 242\).
Time = 0.48 (sec) , antiderivative size = 529, normalized size of antiderivative = 3.75 \[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\left [\frac {2 \, b \sqrt {-\frac {a}{b}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {a}{b}} \cos \left (f x + e\right )}{{\left (a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + b \sqrt {-\frac {a}{b}} \log \left (-\frac {a \cos \left (f x + e\right )^{3} + 4 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {a}{b}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, a \cos \left (f x + e\right )^{2} - 5 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, b^{2} f \sin \left (f x + e\right )}, \frac {2 \, b \sqrt {\frac {a}{b}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {a}{b}} \cos \left (f x + e\right )}{{\left (a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + b \sqrt {\frac {a}{b}} \log \left (\frac {4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {a}{b}} - {\left (a \cos \left (f x + e\right )^{2} + 6 \, a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, b^{2} f \sin \left (f x + e\right )}\right ] \]
[1/4*(2*b*sqrt(-a/b)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos (f*x + e))*sqrt(-a/b)*cos(f*x + e)/((a*cos(f*x + e) + a)*sin(f*x + e)))*si n(f*x + e) + b*sqrt(-a/b)*log(-(a*cos(f*x + e)^3 + 4*sqrt(a*sin(f*x + e))* sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-a/b)*cos(f*x + e)*sin(f*x + e) - 5 *a*cos(f*x + e)^2 - 5*a*cos(f*x + e) + a)/(cos(f*x + e)^3 + 3*cos(f*x + e) ^2 + 3*cos(f*x + e) + 1))*sin(f*x + e) + 8*sqrt(a*sin(f*x + e))*sqrt(b*sin (f*x + e)/cos(f*x + e))*cos(f*x + e))/(b^2*f*sin(f*x + e)), 1/4*(2*b*sqrt( a/b)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt( a/b)*cos(f*x + e)/((a*cos(f*x + e) - a)*sin(f*x + e)))*sin(f*x + e) + b*sq rt(a/b)*log((4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b *sin(f*x + e)/cos(f*x + e))*sqrt(a/b) - (a*cos(f*x + e)^2 + 6*a*cos(f*x + e) + a)*sin(f*x + e))/((cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sin(f*x + e)) )*sin(f*x + e) + 8*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))* cos(f*x + e))/(b^2*f*sin(f*x + e))]
\[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a \sin {\left (e + f x \right )}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {a \sin \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {a \sin \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a\,\sin \left (e+f\,x\right )}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]