Integrand size = 25, antiderivative size = 167 \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}}+\frac {b^{3/2} d \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2} (b \sin (e+f x))^{3/2}}+\frac {b^{3/2} d \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2} (b \sin (e+f x))^{3/2}} \]
-2*d*csc(f*x+e)*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)+b^(3/2)*d*arct an((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/ 2)/(b*sin(f*x+e))^(3/2)+b^(3/2)*d*arctanh((b*sin(f*x+e))^(1/2)/b^(1/2))*(b *tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)/(b*sin(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.70 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.40 \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {5}{4},\frac {9}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} (b \tan (e+f x))^{5/2}}{5 b f \sqrt {d \sec (e+f x)}} \]
(2*Hypergeometric2F1[5/4, 5/4, 9/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4 )*(b*Tan[e + f*x])^(5/2))/(5*b*f*Sqrt[d*Sec[e + f*x]])
Time = 0.39 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.64, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3096, 3042, 3044, 27, 262, 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.
| \(\displaystyle \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}}dx\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle \frac {d (b \tan (e+f x))^{3/2} \int \sec (e+f x) (b \sin (e+f x))^{3/2}dx}{(b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {d (b \tan (e+f x))^{3/2} \int \frac {(b \sin (e+f x))^{3/2}}{\cos (e+f x)}dx}{(b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {d (b \tan (e+f x))^{3/2} \int \frac {b^2 (b \sin (e+f x))^{3/2}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{b f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b d (b \tan (e+f x))^{3/2} \int \frac {(b \sin (e+f x))^{3/2}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {b d (b \tan (e+f x))^{3/2} \left (b^2 \int \frac {1}{\sqrt {b \sin (e+f x)} \left (b^2-b^2 \sin ^2(e+f x)\right )}d(b \sin (e+f x))-2 \sqrt {b \sin (e+f x)}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {b d (b \tan (e+f x))^{3/2} \left (2 b^2 \int \frac {1}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}-2 \sqrt {b \sin (e+f x)}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {b d (b \tan (e+f x))^{3/2} \left (2 b^2 \left (\frac {\int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}}{2 b}\right )-2 \sqrt {b \sin (e+f x)}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {b d (b \tan (e+f x))^{3/2} \left (2 b^2 \left (\frac {\int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 b^{3/2}}\right )-2 \sqrt {b \sin (e+f x)}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b d (b \tan (e+f x))^{3/2} \left (2 b^2 \left (\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 b^{3/2}}\right )-2 \sqrt {b \sin (e+f x)}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
(b*d*(2*b^2*(ArcTan[Sqrt[b]*Sin[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Si n[e + f*x]]/(2*b^(3/2))) - 2*Sqrt[b*Sin[e + f*x]])*(b*Tan[e + f*x])^(3/2)) /(f*(d*Sec[e + f*x])^(3/2)*(b*Sin[e + f*x])^(3/2))
3.4.2.3.1 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] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a^(m + n)*((b*Tan[e + f*x])^n/((a*Sec[e + f*x])^n*(b* Sin[e + f*x])^n)) Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]
Leaf count of result is larger than twice the leaf count of optimal. \(291\) vs. \(2(139)=278\).
Time = 19.50 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.75
| method | result | size |
| default | \(-\frac {\sin \left (f x +e \right ) \left (\operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \cos \left (f x +e \right )+\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )-\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )-2 \sin \left (f x +e \right )\right ) \sqrt {b \tan \left (f x +e \right )}\, b}{f \left (\cos \left (f x +e \right )-1\right ) \sqrt {d \sec \left (f x +e \right )}\, \left (\cos \left (f x +e \right )+1\right )}\) | \(292\) |
-1/f*sin(f*x+e)*(arctanh((sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+c sc(f*x+e)))*(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)-(sin(f*x+e)/(co s(f*x+e)+1)^2)^(1/2)*arctan((sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e )+csc(f*x+e)))*cos(f*x+e)+(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctanh((sin (f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))-(sin(f*x+e)/(cos( f*x+e)+1)^2)^(1/2)*arctan((sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+ csc(f*x+e)))-2*sin(f*x+e))*(b*tan(f*x+e))^(1/2)*b/(cos(f*x+e)-1)/(d*sec(f* x+e))^(1/2)/(cos(f*x+e)+1)
Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (139) = 278\).
Time = 0.55 (sec) , antiderivative size = 741, normalized size of antiderivative = 4.44 \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\left [-\frac {2 \, b d \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b \cos \left (f x + e\right )^{2} - {\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - b\right )}}\right ) - b d \sqrt {-\frac {b}{d}} \log \left (\frac {b \cos \left (f x + e\right )^{4} - 72 \, b \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right ) + 72 \, b}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) + 16 \, b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{8 \, d f}, \frac {2 \, b d \sqrt {\frac {b}{d}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b \cos \left (f x + e\right )^{2} + {\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - b\right )}}\right ) + b d \sqrt {\frac {b}{d}} \log \left (\frac {b \cos \left (f x + e\right )^{4} - 72 \, b \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right ) + 72 \, b}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) - 16 \, b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{8 \, d f}\right ] \]
[-1/8*(2*b*d*sqrt(-b/d)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 - (c os(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqr t(b*sin(f*x + e)/cos(f*x + e))*sqrt(-b/d)*sqrt(d/cos(f*x + e))/(b*cos(f*x + e)^2 - (b*cos(f*x + e) + b)*sin(f*x + e) - b)) - b*d*sqrt(-b/d)*log((b*c os(f*x + e)^4 - 72*b*cos(f*x + e)^2 + 8*(7*cos(f*x + e)^3 - (cos(f*x + e)^ 3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos (f*x + e))*sqrt(-b/d)*sqrt(d/cos(f*x + e)) + 28*(b*cos(f*x + e)^2 - 2*b)*s in(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) + 16*b*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/c os(f*x + e))*cos(f*x + e))/(d*f), 1/8*(2*b*d*sqrt(b/d)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 + (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(b/d)*sq rt(d/cos(f*x + e))/(b*cos(f*x + e)^2 + (b*cos(f*x + e) + b)*sin(f*x + e) - b)) + b*d*sqrt(b/d)*log((b*cos(f*x + e)^4 - 72*b*cos(f*x + e)^2 - 8*(7*co s(f*x + e)^3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(b/d)*sqrt(d/cos(f*x + e)) - 2 8*(b*cos(f*x + e)^2 - 2*b)*sin(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f* x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) - 16*b*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e))/(d*f)]
\[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]
\[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]