Integrand size = 25, antiderivative size = 169 \[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\frac {3 b^{5/2} d \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b^{5/2} d \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f} \]
3/4*b^(5/2)*d*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/f/ (d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/2)-3/4*b^(5/2)*d*arctanh((b*sin(f*x +e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x +e))^(1/2)+1/2*b*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2)/f
Time = 0.74 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.77 \[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\frac {\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \left (3 \arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )-3 \text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+2 \sqrt [4]{\sec ^2(e+f x)} \tan ^{\frac {3}{2}}(e+f x)\right )}{4 f \sqrt [4]{\sec ^2(e+f x)} \tan ^{\frac {5}{2}}(e+f x)} \]
(Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2)*(3*ArcTan[Sqrt[Tan[e + f*x]]/ (Sec[e + f*x]^2)^(1/4)] - 3*ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1 /4)] + 2*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x]^(3/2)))/(4*f*(Sec[e + f*x]^2) ^(1/4)*Tan[e + f*x]^(5/2))
Time = 0.48 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.74, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3091, 3042, 3096, 3042, 3044, 27, 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.
| \(\displaystyle \int (b \tan (e+f x))^{5/2} \sqrt {d \sec (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (b \tan (e+f x))^{5/2} \sqrt {d \sec (e+f x)}dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3}{4} b^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3}{4} b^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^2 d \sqrt {b \tan (e+f x)} \int \sec (e+f x) \sqrt {b \sin (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^2 d \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{\cos (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b d \sqrt {b \tan (e+f x)} \int \frac {b^2 \sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^3 d \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^3 d \sqrt {b \tan (e+f x)} \int \frac {b^2 \sin ^2(e+f x)}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^3 d \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^3 d \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f}-\frac {3 b^3 d \sqrt {b \tan (e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
(-3*b^3*d*(-1/2*ArcTan[Sqrt[b]*Sin[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sin [e + f*x]]/(2*Sqrt[b]))*Sqrt[b*Tan[e + f*x]])/(2*f*Sqrt[d*Sec[e + f*x]]*Sq rt[b*Sin[e + f*x]]) + (b*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(2*f )
3.4.9.3.1 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_)]^(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[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*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]
Time = 12.08 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.38
| method | result | size |
| default | \(-\frac {\sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, b^{2} \left (2 \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\sin ^{3}\left (f x +e \right )\right )-3 \cos \left (f x +e \right ) \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )-3 \cos \left (f x +e \right ) \operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \left (\sin ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{4 f \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(233\) |
-1/4/f*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)*b^2/(cos(f*x+e)-1)/(cos(f *x+e)+1)^2/(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(2*(sin(f*x+e)/(cos(f*x+e)+ 1)^2)^(1/2)*sin(f*x+e)^3-3*cos(f*x+e)*arctan((sin(f*x+e)/(cos(f*x+e)+1)^2) ^(1/2)*(cot(f*x+e)+csc(f*x+e)))*sin(f*x+e)^2-3*cos(f*x+e)*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)^2+2*sin(f* x+e)^2*tan(f*x+e)*(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (135) = 270\).
Time = 0.39 (sec) , antiderivative size = 788, normalized size of antiderivative = 4.66 \[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\left [\frac {6 \, \sqrt {-b d} b^{2} \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 {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d - {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) \cos \left (f x + e\right ) + 3 \, \sqrt {-b d} b^{2} \cos \left (f x + e\right ) \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b d \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 {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d + 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\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^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{32 \, f \cos \left (f x + e\right )}, \frac {6 \, \sqrt {b d} b^{2} \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 {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d + {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) \cos \left (f x + e\right ) + 3 \, \sqrt {b d} b^{2} \cos \left (f x + e\right ) \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b d \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 {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d - 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\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^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{32 \, f \cos \left (f x + e\right )}\right ] \]
[1/32*(6*sqrt(-b*d)*b^2*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(f* x + e)^2 - b*d - (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e) + 3* sqrt(-b*d)*b^2*cos(f*x + e)*log((b*d*cos(f*x + e)^4 - 72*b*d*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f *x + e)) + 72*b*d + 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) + 1 6*b^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x + e)) /(f*cos(f*x + e)), 1/32*(6*sqrt(b*d)*b^2*arctan(1/4*(cos(f*x + e)^3 - 5*co s(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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(f*x + e)^2 - b*d + (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*c os(f*x + e) + 3*sqrt(b*d)*b^2*cos(f*x + e)*log((b*d*cos(f*x + e)^4 - 72*b* d*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e) )*sqrt(d/cos(f*x + e)) + 72*b*d - 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(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^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + ...
Timed out. \[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\int { \sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx=\int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]