Integrand size = 25, antiderivative size = 145 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {b^2 \arctan \left (\sqrt {\cos (e+f x)}\right ) \sqrt {a \sin (e+f x)}}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {b^2 \text {arctanh}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {a \sin (e+f x)}}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}} \]
b^2*arctan(cos(f*x+e)^(1/2))*(a*sin(f*x+e))^(1/2)/a^3/f/cos(f*x+e)^(1/2)/( b*tan(f*x+e))^(1/2)-b^2*arctanh(cos(f*x+e)^(1/2))*(a*sin(f*x+e))^(1/2)/a^3 /f/cos(f*x+e)^(1/2)/(b*tan(f*x+e))^(1/2)+2*b*(b*tan(f*x+e))^(1/2)/a^2/f/(a *sin(f*x+e))^(1/2)
Time = 0.62 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.72 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {b \left (\arctan \left (\sqrt [4]{\cos ^2(e+f x)}\right ) \cos ^2(e+f x)-\text {arctanh}\left (\sqrt [4]{\cos ^2(e+f x)}\right ) \cos ^2(e+f x)+2 \cos ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{a^2 f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)}} \]
(b*(ArcTan[(Cos[e + f*x]^2)^(1/4)]*Cos[e + f*x]^2 - ArcTanh[(Cos[e + f*x]^ 2)^(1/4)]*Cos[e + f*x]^2 + 2*(Cos[e + f*x]^2)^(3/4))*Sqrt[b*Tan[e + f*x]]) /(a^2*f*(Cos[e + f*x]^2)^(3/4)*Sqrt[a*Sin[e + f*x]])
Time = 0.47 (sec) , antiderivative size = 110, 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, 3073, 3042, 3081, 27, 3042, 3045, 266, 827, 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.
| \(\displaystyle \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3073 |
| \(\displaystyle \frac {b^2 \int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}dx}{a^2}+\frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}dx}{a^2}+\frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3081 |
| \(\displaystyle \frac {b^2 \sqrt {a \sin (e+f x)} \int \frac {\sqrt {\cos (e+f x)} \csc (e+f x)}{a}dx}{a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \sqrt {a \sin (e+f x)} \int \sqrt {\cos (e+f x)} \csc (e+f x)dx}{a^3 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {a \sin (e+f x)} \int \frac {\sqrt {\cos (e+f x)}}{\sin (e+f x)}dx}{a^3 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle \frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {b^2 \sqrt {a \sin (e+f x)} \int \frac {\sqrt {\cos (e+f x)}}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b^2 \sqrt {a \sin (e+f x)} \int \frac {\cos (e+f x)}{1-\cos ^2(e+f x)}d\sqrt {\cos (e+f x)}}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b^2 \sqrt {a \sin (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}-\frac {1}{2} \int \frac {1}{\cos (e+f x)+1}d\sqrt {\cos (e+f x)}\right )}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b^2 \sqrt {a \sin (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}-\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )\right )}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 b \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b^2 \sqrt {a \sin (e+f x)} \left (\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (e+f x)}\right )-\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )\right )}{a^3 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
(-2*b^2*(-1/2*ArcTan[Sqrt[Cos[e + f*x]]] + ArcTanh[Sqrt[Cos[e + f*x]]]/2)* Sqrt[a*Sin[e + f*x]])/(a^3*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (2 *b*Sqrt[b*Tan[e + f*x]])/(a^2*f*Sqrt[a*Sin[e + f*x]])
3.2.25.3.1 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) 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[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1)/ (a^2*f*(n - 1))), x] - Simp[b^2*((m + 2)/(a^2*(n - 1))) Int[(a*Sin[e + f* x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && G tQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2])) && IntegersQ[2*m, 2 *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 = 1.99 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.52
| method | result | size |
| default | \(\frac {\left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+\cos \left (f x +e \right ) \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \sqrt {b \tan \left (f x +e \right )}\, b}{2 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {\sin \left (f x +e \right ) a}\, a^{2}}\) | \(221\) |
1/2/f*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+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*(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))+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*(b* tan(f*x+e))^(1/2)*b/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(s in(f*x+e)*a)^(1/2)/a^2
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (125) = 250\).
Time = 0.48 (sec) , antiderivative size = 524, normalized size of antiderivative = 3.61 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\left [\frac {2 \, a b \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 ) \sin \left (f x + e\right ) + a b \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 ) \sin \left (f x + e\right ) + 8 \, \sqrt {a \sin \left (f x + e\right )} b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{4 \, a^{3} f \sin \left (f x + e\right )}, -\frac {2 \, a b \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 ) \sin \left (f x + e\right ) - a b \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 ) \sin \left (f x + e\right ) - 8 \, \sqrt {a \sin \left (f x + e\right )} b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{4 \, a^{3} f \sin \left (f x + e\right )}\right ] \]
[1/4*(2*a*b*sqrt(-b/a)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/c os(f*x + e))*sqrt(-b/a)*cos(f*x + e)/((b*cos(f*x + e) + b)*sin(f*x + e)))* sin(f*x + e) + a*b*sqrt(-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))*sin(f*x + e) + 8*sqrt(a*sin(f*x + e))*b*sqrt (b*sin(f*x + e)/cos(f*x + e)))/(a^3*f*sin(f*x + e)), -1/4*(2*a*b*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)))*sin(f*x + e) - a*b*sqrt (b/a)*log((4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*s in(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)))* sin(f*x + e) - 8*sqrt(a*sin(f*x + e))*b*sqrt(b*sin(f*x + e)/cos(f*x + e))) /(a^3*f*sin(f*x + e))]
Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]