Integrand size = 25, antiderivative size = 107 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=-\frac {\arctan \left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}}-\frac {\text {arctanh}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}} \]
-arctan(cos(f*x+e)^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/a/f/(a*sin (f*x+e))^(1/2)-arctanh(cos(f*x+e)^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^( 1/2)/a/f/(a*sin(f*x+e))^(1/2)
Time = 0.57 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=-\frac {b \left (\arctan \left (\sqrt [4]{\cos ^2(e+f x)}\right )+\text {arctanh}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \]
-((b*(ArcTan[(Cos[e + f*x]^2)^(1/4)] + ArcTanh[(Cos[e + f*x]^2)^(1/4)])*Sq rt[a*Sin[e + f*x]])/(a^2*f*(Cos[e + f*x]^2)^(1/4)*Sqrt[b*Tan[e + f*x]]))
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{a \sqrt {\cos (e+f x)}}dx}{\sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{\sqrt {\cos (e+f x)}}dx}{a \sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \sin (e+f x)}dx}{a \sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (1-\cos ^2(e+f x)\right )}d\cos (e+f x)}{a f \sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{1-\cos ^2(e+f x)}d\sqrt {\cos (e+f x)}}{a f \sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \int \frac {1}{\cos (e+f x)+1}d\sqrt {\cos (e+f x)}\right )}{a f \sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )\right )}{a f \sqrt {a \sin (e+f x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (e+f x)}\right )\right )}{a f \sqrt {a \sin (e+f x)}}\) |
(-2*(ArcTan[Sqrt[Cos[e + f*x]]]/2 + ArcTanh[Sqrt[Cos[e + f*x]]]/2)*Sqrt[Co s[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(a*f*Sqrt[a*Sin[e + f*x]])
3.2.18.3.1 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[((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 = 1.39 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53
method | result | size |
default | \(\frac {\cos \left (f x +e \right ) \left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )\right ) \sqrt {b \tan \left (f x +e \right )}}{2 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {\sin \left (f x +e \right ) a}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a}\) | \(164\) |
1/2/f*cos(f*x+e)*(arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-ln(2*(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)))*(b*tan(f*x+e))^(1/2)/(cos(f*x +e)+1)/(sin(f*x+e)*a)^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/a
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (91) = 182\).
Time = 0.42 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.86 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\left [\frac {2 \, \sqrt {-\frac {b}{a}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}\right ) + \sqrt {-\frac {b}{a}} \log \left (-\frac {b \cos \left (f x + e\right )^{3} + 4 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, b \cos \left (f x + e\right )^{2} - 5 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, a f}, \frac {2 \, \sqrt {\frac {b}{a}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) - b\right )} \sin \left (f x + e\right )}\right ) + \sqrt {\frac {b}{a}} \log \left (\frac {4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{a}} - {\left (b \cos \left (f x + e\right )^{2} + 6 \, b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right )}{4 \, a f}\right ] \]
[1/4*(2*sqrt(-b/a)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f *x + e))*sqrt(-b/a)*cos(f*x + e)/((b*cos(f*x + e) + b)*sin(f*x + e))) + sq rt(-b/a)*log(-(b*cos(f*x + e)^3 + 4*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-b/a)*cos(f*x + e)*sin(f*x + e) - 5*b*cos(f*x + e)^2 - 5*b*cos(f*x + e) + b)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x + e) + 1)))/(a*f), 1/4*(2*sqrt(b/a)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin (f*x + e)/cos(f*x + e))*sqrt(b/a)*cos(f*x + e)/((b*cos(f*x + e) - b)*sin(f *x + e))) + sqrt(b/a)*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(b/a) - (b*cos(f*x + e)^2 + 6*b*cos(f*x + e) + b)*sin(f*x + e))/((cos(f*x + e)^2 - 2*cos(f*x + e) + 1) *sin(f*x + e))))/(a*f)]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {b \tan {\left (e + f x \right )}}}{\left (a \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]