Integrand size = 25, antiderivative size = 151 \[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}+\frac {\arctan \left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{4 a b^2 f \sqrt {a \sin (e+f x)}}+\frac {\text {arctanh}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{4 a b^2 f \sqrt {a \sin (e+f x)}} \]
-1/2/b/f/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2)+1/4*arctan(cos(f*x+e)^( 1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/a/b^2/f/(a*sin(f*x+e))^(1/2)+1 /4*arctanh(cos(f*x+e)^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/a/b^2/f /(a*sin(f*x+e))^(1/2)
Time = 0.60 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=\frac {\left (\arctan \left (\sqrt [4]{\cos ^2(e+f x)}\right )+\text {arctanh}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-2 \sqrt [4]{\cos ^2(e+f x)} \csc ^2(e+f x)\right ) \sin ^2(e+f x)}{4 b f \sqrt [4]{\cos ^2(e+f x)} (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \]
((ArcTan[(Cos[e + f*x]^2)^(1/4)] + ArcTanh[(Cos[e + f*x]^2)^(1/4)] - 2*(Co s[e + f*x]^2)^(1/4)*Csc[e + f*x]^2)*Sin[e + f*x]^2)/(4*b*f*(Cos[e + f*x]^2 )^(1/4)*(a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])
Time = 0.47 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.75, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3077, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3077 |
| \(\displaystyle -\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}}dx}{4 b^2}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}}dx}{4 b^2}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3081 |
| \(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{a \sqrt {\cos (e+f x)}}dx}{4 b^2 \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{\sqrt {\cos (e+f x)}}dx}{4 a b^2 \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \sin (e+f x)}dx}{4 a b^2 \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (1-\cos ^2(e+f x)\right )}d\cos (e+f x)}{4 a b^2 f \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{1-\cos ^2(e+f x)}d\sqrt {\cos (e+f x)}}{2 a b^2 f \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \int \frac {1}{\cos (e+f x)+1}d\sqrt {\cos (e+f x)}\right )}{2 a b^2 f \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )\right )}{2 a b^2 f \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (e+f x)}\right )\right )}{2 a b^2 f \sqrt {a \sin (e+f x)}}-\frac {1}{2 b f (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}\) |
-1/2*1/(b*f*(a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]) + ((ArcTan[Sqrt[C os[e + f*x]]]/2 + ArcTanh[Sqrt[Cos[e + f*x]]]/2)*Sqrt[Cos[e + f*x]]*Sqrt[b *Tan[e + f*x]])/(2*a*b^2*f*Sqrt[a*Sin[e + f*x]])
3.2.38.3.1 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[(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n + 1))), x] - Simp[(n + 1)/(b^2*(m + n + 1)) Int[(a*Sin[e + f*x])^m*( b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n] && !(EqQ[n, -3/2] && EqQ[m, 1] )
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])
Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(125)=250\).
Time = 1.10 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.91
| method | result | size |
| default | \(\frac {\csc \left (f x +e \right ) \left (\cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\cos \left (f x +e \right ) \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )-\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )-4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{8 f \sqrt {b \tan \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) a}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a b}\) | \(288\) |
1/8/f*csc(f*x+e)*(cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/ 2))-cos(f*x+e)*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-c os(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))-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)/(c os(f*x+e)+1))-4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))/(b*tan(f*x+e))^(1/2) /(sin(f*x+e)*a)^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/a/b
Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (125) = 250\).
Time = 0.47 (sec) , antiderivative size = 608, normalized size of antiderivative = 4.03 \[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=\left [-\frac {2 \, \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {2 \, \sqrt {-a b} \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 )}{{\left (a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )^{2} + 4 \, \sqrt {-a b} \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 ) \sin \left (f x + e\right ) - 5 \, a b \cos \left (f x + e\right ) + a 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 ) \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 )}{16 \, {\left (a^{2} b^{2} f \cos \left (f x + e\right )^{2} - a^{2} b^{2} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {2 \, \sqrt {a b} \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 )}{{\left (a b \cos \left (f x + e\right ) - a b\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {4 \, \sqrt {a b} {\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 )}} + {\left (a b \cos \left (f x + e\right )^{2} + 6 \, a b \cos \left (f x + e\right ) + a 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 ) \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 )}{16 \, {\left (a^{2} b^{2} f \cos \left (f x + e\right )^{2} - a^{2} b^{2} f\right )} \sin \left (f x + e\right )}\right ] \]
[-1/16*(2*sqrt(-a*b)*(cos(f*x + e)^2 - 1)*arctan(2*sqrt(-a*b)*sqrt(a*sin(f *x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f*x + e) + a*b)*sin(f*x + e)))*sin(f*x + e) + sqrt(-a*b)*(cos(f*x + e)^2 - 1)*log( -(a*b*cos(f*x + e)^3 - 5*a*b*cos(f*x + e)^2 + 4*sqrt(-a*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 5*a*b* cos(f*x + e) + a*b)/(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))/((a^2*b^2*f*cos(f*x + e)^2 - a^2*b^2*f)*sin(f*x + e)), -1/ 16*(2*sqrt(a*b)*(cos(f*x + e)^2 - 1)*arctan(2*sqrt(a*b)*sqrt(a*sin(f*x + e ))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f*x + e) - a*b )*sin(f*x + e)))*sin(f*x + e) - sqrt(a*b)*(cos(f*x + e)^2 - 1)*log(-(4*sqr t(a*b)*(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)) + (a*b*cos(f*x + e)^2 + 6*a*b*cos(f*x + e) + a*b)*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) )/((a^2*b^2*f*cos(f*x + e)^2 - a^2*b^2*f)*sin(f*x + e))]
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]