Integrand size = 25, antiderivative size = 178 \[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=-\frac {\sqrt {b} d^3 \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {\sqrt {b} d^3 \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {d^2 \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f} \] Output:
-1/4*b^(1/2)*d^3*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)+1/4*b^(1/2)*d^3*arctanh((b*si n(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(1/2)/(b*si n(f*x+e))^(1/2)+1/2*d^2*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2)/b/f
Time = 0.77 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.74 \[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\frac {d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \left (-\arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+2 \sqrt [4]{\sec ^2(e+f x)} \tan ^{\frac {3}{2}}(e+f x)\right )}{4 f \sqrt [4]{\sec ^2(e+f x)} \sqrt {\tan (e+f x)}} \] Input:
Integrate[(d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]
Output:
(d^2*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]]*(-ArcTan[Sqrt[Tan[e + f*x]] /(Sec[e + f*x]^2)^(1/4)] + 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)*Sqrt[Tan[e + f*x]])
Time = 0.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3093, 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.
\(\displaystyle \int \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}dx\) |
\(\Big \downarrow \) 3093 |
\(\displaystyle \frac {1}{4} d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 3096 |
\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \sec (e+f x) \sqrt {b \sin (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{\cos (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {b d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sin ^2(e+f x)}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\) |
Input:
Int[(d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]
Output:
(b*d^3*(-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]]*Sqrt[ b*Sin[e + f*x]]) + (d^2*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(2*b* f)
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[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1)) Int[(a*Sec[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && 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 = 5.69 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {\sqrt {b \tan \left (f x +e \right )}\, d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (-\cos \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )-\cos \left (f x +e \right ) \arctan \left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )+\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (2 \sin \left (f x +e \right )+2 \tan \left (f x +e \right )\right )\right )}{4 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\) | \(186\) |
Input:
int((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4/f*(b*tan(f*x+e))^(1/2)*d^2*(d*sec(f*x+e))^(1/2)/(1+cos(f*x+e))/(sin(f* x+e)/(1+cos(f*x+e))^2)^(1/2)*(-cos(f*x+e)*arctanh((sin(f*x+e)/(1+cos(f*x+e ))^2)^(1/2)*sin(f*x+e)/(cos(f*x+e)-1))-cos(f*x+e)*arctan((sin(f*x+e)/(1+co s(f*x+e))^2)^(1/2)*sin(f*x+e)/(cos(f*x+e)-1))+(sin(f*x+e)/(1+cos(f*x+e))^2 )^(1/2)*(2*sin(f*x+e)+2*tan(f*x+e)))
Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (144) = 288\).
Time = 0.29 (sec) , antiderivative size = 788, normalized size of antiderivative = 4.43 \[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx =\text {Too large to display} \] Input:
integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
[-1/32*(2*sqrt(-b*d)*d^2*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)*sq rt(-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) - s qrt(-b*d)*d^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 *d^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*(2*sqrt(b*d)*d^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) - sqrt(b*d)*d^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))*s in(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*d^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)...
Timed out. \[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\text {Timed out} \] Input:
integrate((d*sec(f*x+e))**(5/2)*(b*tan(f*x+e))**(1/2),x)
Output:
Timed out
\[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )} \,d x } \] Input:
integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)
\[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )} \,d x } \] Input:
integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)
Timed out. \[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\int \sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \] Input:
int((b*tan(e + f*x))^(1/2)*(d/cos(e + f*x))^(5/2),x)
Output:
int((b*tan(e + f*x))^(1/2)*(d/cos(e + f*x))^(5/2), x)
\[ \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\sqrt {d}\, \sqrt {b}\, \left (\int \sqrt {\tan \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}d x \right ) d^{2} \] Input:
int((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x)
Output:
sqrt(d)*sqrt(b)*int(sqrt(tan(e + f*x))*sqrt(sec(e + f*x))*sec(e + f*x)**2, x)*d**2