Integrand size = 25, antiderivative size = 208 \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\frac {3 b^{5/2} d^3 \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{32 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b^{5/2} d^3 \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{32 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b d^2 \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac {b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f} \]
3/32*b^(5/2)*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)-3/32*b^(5/2)*d^3*arctanh((b*s in(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(1/2)/(b*s in(f*x+e))^(1/2)+1/4*b*(d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2)/f-3/16*b* d^2*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2)/f
Time = 2.33 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.74 \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\frac {d^4 (b \tan (e+f x))^{5/2} \left (3 \arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right ) \sec ^2(e+f x)^{3/4}-3 \text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right ) \sec ^2(e+f x)^{3/4}+2 \sec ^2(e+f x) \left (-3+4 \sec ^2(e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)\right )}{32 f (d \sec (e+f x))^{3/2} \tan ^{\frac {5}{2}}(e+f x)} \]
(d^4*(b*Tan[e + f*x])^(5/2)*(3*ArcTan[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^ (1/4)]*(Sec[e + f*x]^2)^(3/4) - 3*ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x] ^2)^(1/4)]*(Sec[e + f*x]^2)^(3/4) + 2*Sec[e + f*x]^2*(-3 + 4*Sec[e + f*x]^ 2)*Tan[e + f*x]^(3/2)))/(32*f*(d*Sec[e + f*x])^(3/2)*Tan[e + f*x]^(5/2))
Time = 0.67 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.82, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3091, 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.
| \(\displaystyle \int (b \tan (e+f x))^{5/2} (d \sec (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (b \tan (e+f x))^{5/2} (d \sec (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 3093 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {1}{4} d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {1}{4} d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {d^3 \sqrt {b \tan (e+f x)} \int \sec (e+f x) \sqrt {b \sin (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{\cos (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sin ^2(e+f x)}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\) |
(b*(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2))/(4*f) - (3*b^2*((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)))/8
3.4.7.3.1 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^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 = 169.05 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.44
| method | result | size |
| default | \(\frac {\sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, b^{2} d^{2} \left (3 \cos \left (f x +e \right ) \operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )+3 \cos \left (f x +e \right ) \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )+6 \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\sin ^{3}\left (f x +e \right )\right )+6 \left (\sin ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-8 \sin \left (f x +e \right ) \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-8 \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{32 f \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(300\) |
1/32/f*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)*b^2*d^2/(cos(f*x+e)-1)/(c os(f*x+e)+1)^2/(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(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+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+6*(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)^3+6*si n(f*x+e)^2*tan(f*x+e)*(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-8*sin(f*x+e)*tan (f*x+e)^2*(sin(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-8*tan(f*x+e)^3*(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. 422 vs. \(2 (168) = 336\).
Time = 0.45 (sec) , antiderivative size = 852, normalized size of antiderivative = 4.10 \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
[1/256*(6*sqrt(-b*d)*b^2*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 )*sqrt(-b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*c os(f*x + e)^2 - b*d - (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e) ^3 + 3*sqrt(-b*d)*b^2*d^2*cos(f*x + e)^3*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*(3*b^2*d^2*cos(f*x + e)^2 - 4*b^2*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)^3), 1/256 *(6*sqrt(b*d)*b^2*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)*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)))*cos(f*x + e)^3 + 3*s qrt(b*d)*b^2*d^2*cos(f*x + e)^3*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/co s(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...
Timed out. \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \]