Integrand size = 14, antiderivative size = 110 \[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\frac {b \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-b x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}-\frac {b \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f} \]
b*cot(f*x+e)*(b*tan(f*x+e)^4)^(1/2)/f-b*x*cot(f*x+e)^2*(b*tan(f*x+e)^4)^(1 /2)-1/3*b*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)/f+1/5*b*(b*tan(f*x+e)^4)^(1/2) *tan(f*x+e)^3/f
Time = 0.91 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60 \[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\frac {\cot (e+f x) \left (3-5 \cot ^2(e+f x)+15 \cot ^4(e+f x)-15 \arctan (\tan (e+f x)) \cot ^5(e+f x)\right ) \left (b \tan ^4(e+f x)\right )^{3/2}}{15 f} \]
(Cot[e + f*x]*(3 - 5*Cot[e + f*x]^2 + 15*Cot[e + f*x]^4 - 15*ArcTan[Tan[e + f*x]]*Cot[e + f*x]^5)*(b*Tan[e + f*x]^4)^(3/2))/(15*f)
Time = 0.38 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 3042, 3954, 24}
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.
\(\displaystyle \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan (e+f x)^4\right )^{3/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \int \tan ^6(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \int \tan (e+f x)^6dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\frac {\tan ^5(e+f x)}{5 f}-\int \tan ^4(e+f x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\frac {\tan ^5(e+f x)}{5 f}-\int \tan (e+f x)^4dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\int \tan ^2(e+f x)dx+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\int \tan (e+f x)^2dx+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (-\int 1dx+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}+\frac {\tan (e+f x)}{f}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b \left (\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}+\frac {\tan (e+f x)}{f}-x\right ) \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}\) |
b*Cot[e + f*x]^2*Sqrt[b*Tan[e + f*x]^4]*(-x + Tan[e + f*x]/f - Tan[e + f*x ]^3/(3*f) + Tan[e + f*x]^5/(5*f))
3.1.14.3.1 Defintions of rubi rules used
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58
method | result | size |
derivativedivides | \(-\frac {\left (b \tan \left (f x +e \right )^{4}\right )^{\frac {3}{2}} \left (-3 \tan \left (f x +e \right )^{5}+5 \tan \left (f x +e \right )^{3}+15 \arctan \left (\tan \left (f x +e \right )\right )-15 \tan \left (f x +e \right )\right )}{15 f \tan \left (f x +e \right )^{6}}\) | \(64\) |
default | \(-\frac {\left (b \tan \left (f x +e \right )^{4}\right )^{\frac {3}{2}} \left (-3 \tan \left (f x +e \right )^{5}+5 \tan \left (f x +e \right )^{3}+15 \arctan \left (\tan \left (f x +e \right )\right )-15 \tan \left (f x +e \right )\right )}{15 f \tan \left (f x +e \right )^{6}}\) | \(64\) |
risch | \(\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {2 i b \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left (45 \,{\mathrm e}^{8 i \left (f x +e \right )}+90 \,{\mathrm e}^{6 i \left (f x +e \right )}+140 \,{\mathrm e}^{4 i \left (f x +e \right )}+70 \,{\mathrm e}^{2 i \left (f x +e \right )}+23\right )}{15 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} f}\) | \(170\) |
-1/15/f*(b*tan(f*x+e)^4)^(3/2)*(-3*tan(f*x+e)^5+5*tan(f*x+e)^3+15*arctan(t an(f*x+e))-15*tan(f*x+e))/tan(f*x+e)^6
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.56 \[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\frac {{\left (3 \, b \tan \left (f x + e\right )^{5} - 5 \, b \tan \left (f x + e\right )^{3} - 15 \, b f x + 15 \, b \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{15 \, f \tan \left (f x + e\right )^{2}} \]
1/15*(3*b*tan(f*x + e)^5 - 5*b*tan(f*x + e)^3 - 15*b*f*x + 15*b*tan(f*x + e))*sqrt(b*tan(f*x + e)^4)/(f*tan(f*x + e)^2)
\[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\int \left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.48 \[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\frac {3 \, b^{\frac {3}{2}} \tan \left (f x + e\right )^{5} - 5 \, b^{\frac {3}{2}} \tan \left (f x + e\right )^{3} - 15 \, {\left (f x + e\right )} b^{\frac {3}{2}} + 15 \, b^{\frac {3}{2}} \tan \left (f x + e\right )}{15 \, f} \]
1/15*(3*b^(3/2)*tan(f*x + e)^5 - 5*b^(3/2)*tan(f*x + e)^3 - 15*(f*x + e)*b ^(3/2) + 15*b^(3/2)*tan(f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 992 vs. \(2 (98) = 196\).
Time = 2.41 (sec) , antiderivative size = 992, normalized size of antiderivative = 9.02 \[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
1/60*(15*pi - 60*f*x*tan(f*x)^5*tan(e)^5 - 15*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*tan(f*x)^5*tan(e)^5 - 15*pi* tan(f*x)^5*tan(e)^5 + 30*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e))) *tan(f*x)^5*tan(e)^5 + 30*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1) )*tan(f*x)^5*tan(e)^5 + 300*f*x*tan(f*x)^4*tan(e)^4 + 75*pi*sgn(2*tan(f*x) ^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*tan(f*x)^4*tan(e) ^4 + 75*pi*tan(f*x)^4*tan(e)^4 - 150*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x ) + tan(e)))*tan(f*x)^4*tan(e)^4 - 150*arctan((tan(f*x) + tan(e))/(tan(f*x )*tan(e) - 1))*tan(f*x)^4*tan(e)^4 - 60*tan(f*x)^5*tan(e)^4 - 60*tan(f*x)^ 4*tan(e)^5 - 600*f*x*tan(f*x)^3*tan(e)^3 - 150*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*tan(f*x)^3*tan(e)^3 + 20*ta n(f*x)^5*tan(e)^2 - 150*pi*tan(f*x)^3*tan(e)^3 + 300*arctan((tan(f*x)*tan( e) - 1)/(tan(f*x) + tan(e)))*tan(f*x)^3*tan(e)^3 + 300*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)^3*tan(e)^3 + 300*tan(f*x)^4*tan(e) ^3 + 300*tan(f*x)^3*tan(e)^4 + 20*tan(f*x)^2*tan(e)^5 + 600*f*x*tan(f*x)^2 *tan(e)^2 + 150*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f *x) - 2*tan(e))*tan(f*x)^2*tan(e)^2 - 12*tan(f*x)^5 - 100*tan(f*x)^4*tan(e ) + 150*pi*tan(f*x)^2*tan(e)^2 - 300*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x ) + tan(e)))*tan(f*x)^2*tan(e)^2 - 300*arctan((tan(f*x) + tan(e))/(tan(f*x )*tan(e) - 1))*tan(f*x)^2*tan(e)^2 - 600*tan(f*x)^3*tan(e)^2 - 600*tan(...
Timed out. \[ \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{3/2} \,d x \]