Integrand size = 14, antiderivative size = 182 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {b^2 \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-b^2 x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-\frac {b^2 \tan ^5(e+f x) \sqrt {b \tan ^4(e+f x)}}{7 f}+\frac {b^2 \tan ^7(e+f x) \sqrt {b \tan ^4(e+f x)}}{9 f} \]
b^2*cot(f*x+e)*(b*tan(f*x+e)^4)^(1/2)/f-b^2*x*cot(f*x+e)^2*(b*tan(f*x+e)^4 )^(1/2)-1/3*b^2*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)/f+1/5*b^2*(b*tan(f*x+e)^ 4)^(1/2)*tan(f*x+e)^3/f-1/7*b^2*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)^5/f+1/9* b^2*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)^7/f
Time = 0.91 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.47 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {\cot (e+f x) \left (35-45 \cot ^2(e+f x)+63 \cot ^4(e+f x)-105 \cot ^6(e+f x)+315 \cot ^8(e+f x)-315 \arctan (\tan (e+f x)) \cot ^9(e+f x)\right ) \left (b \tan ^4(e+f x)\right )^{5/2}}{315 f} \]
(Cot[e + f*x]*(35 - 45*Cot[e + f*x]^2 + 63*Cot[e + f*x]^4 - 105*Cot[e + f* x]^6 + 315*Cot[e + f*x]^8 - 315*ArcTan[Tan[e + f*x]]*Cot[e + f*x]^9)*(b*Ta n[e + f*x]^4)^(5/2))/(315*f)
Time = 0.56 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.55, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 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.
\(\displaystyle \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan (e+f x)^4\right )^{5/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \int \tan ^{10}(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \int \tan (e+f x)^{10}dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\frac {\tan ^9(e+f x)}{9 f}-\int \tan ^8(e+f x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\frac {\tan ^9(e+f x)}{9 f}-\int \tan (e+f x)^8dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\int \tan ^6(e+f x)dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\int \tan (e+f x)^6dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (-\int \tan ^4(e+f x)dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}+\frac {\tan ^5(e+f x)}{5 f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (-\int \tan (e+f x)^4dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}+\frac {\tan ^5(e+f x)}{5 f}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\int \tan ^2(e+f x)dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (\int \tan (e+f x)^2dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \left (-\int 1dx+\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}+\frac {\tan (e+f x)}{f}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b^2 \left (\frac {\tan ^9(e+f x)}{9 f}-\frac {\tan ^7(e+f x)}{7 f}+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}+\frac {\tan (e+f x)}{f}-x\right ) \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}\) |
b^2*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) - Tan[e + f*x]^7/(7*f) + Tan[e + f*x]^9 /(9*f))
3.1.13.3.1 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 = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.46
method | result | size |
derivativedivides | \(-\frac {\left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}} \left (-35 \tan \left (f x +e \right )^{9}+45 \tan \left (f x +e \right )^{7}-63 \tan \left (f x +e \right )^{5}+105 \tan \left (f x +e \right )^{3}+315 \arctan \left (\tan \left (f x +e \right )\right )-315 \tan \left (f x +e \right )\right )}{315 f \tan \left (f x +e \right )^{10}}\) | \(84\) |
default | \(-\frac {\left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}} \left (-35 \tan \left (f x +e \right )^{9}+45 \tan \left (f x +e \right )^{7}-63 \tan \left (f x +e \right )^{5}+105 \tan \left (f x +e \right )^{3}+315 \arctan \left (\tan \left (f x +e \right )\right )-315 \tan \left (f x +e \right )\right )}{315 f \tan \left (f x +e \right )^{10}}\) | \(84\) |
risch | \(\frac {b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {2 i b^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left (1575 \,{\mathrm e}^{16 i \left (f x +e \right )}+6300 \,{\mathrm e}^{14 i \left (f x +e \right )}+21000 \,{\mathrm e}^{12 i \left (f x +e \right )}+31500 \,{\mathrm e}^{10 i \left (f x +e \right )}+39438 \,{\mathrm e}^{8 i \left (f x +e \right )}+26292 \,{\mathrm e}^{6 i \left (f x +e \right )}+13968 \,{\mathrm e}^{4 i \left (f x +e \right )}+3492 \,{\mathrm e}^{2 i \left (f x +e \right )}+563\right )}{315 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7} f}\) | \(218\) |
-1/315/f*(b*tan(f*x+e)^4)^(5/2)*(-35*tan(f*x+e)^9+45*tan(f*x+e)^7-63*tan(f *x+e)^5+105*tan(f*x+e)^3+315*arctan(tan(f*x+e))-315*tan(f*x+e))/tan(f*x+e) ^10
Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.53 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {{\left (35 \, b^{2} \tan \left (f x + e\right )^{9} - 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, b^{2} \tan \left (f x + e\right )^{5} - 105 \, b^{2} \tan \left (f x + e\right )^{3} - 315 \, b^{2} f x + 315 \, b^{2} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{315 \, f \tan \left (f x + e\right )^{2}} \]
1/315*(35*b^2*tan(f*x + e)^9 - 45*b^2*tan(f*x + e)^7 + 63*b^2*tan(f*x + e) ^5 - 105*b^2*tan(f*x + e)^3 - 315*b^2*f*x + 315*b^2*tan(f*x + e))*sqrt(b*t an(f*x + e)^4)/(f*tan(f*x + e)^2)
\[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\int \left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
Time = 0.35 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.43 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {35 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{9} - 45 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{7} + 63 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{5} - 105 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} b^{\frac {5}{2}} + 315 \, b^{\frac {5}{2}} \tan \left (f x + e\right )}{315 \, f} \]
1/315*(35*b^(5/2)*tan(f*x + e)^9 - 45*b^(5/2)*tan(f*x + e)^7 + 63*b^(5/2)* tan(f*x + e)^5 - 105*b^(5/2)*tan(f*x + e)^3 - 315*(f*x + e)*b^(5/2) + 315* b^(5/2)*tan(f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 960 vs. \(2 (162) = 324\).
Time = 5.88 (sec) , antiderivative size = 960, normalized size of antiderivative = 5.27 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\text {Too large to display} \]
-1/315*(315*b^2*f*x*tan(f*x)^9*tan(e)^9 - 2835*b^2*f*x*tan(f*x)^8*tan(e)^8 + 315*b^2*tan(f*x)^9*tan(e)^8 + 315*b^2*tan(f*x)^8*tan(e)^9 + 11340*b^2*f *x*tan(f*x)^7*tan(e)^7 - 105*b^2*tan(f*x)^9*tan(e)^6 - 2835*b^2*tan(f*x)^8 *tan(e)^7 - 2835*b^2*tan(f*x)^7*tan(e)^8 - 105*b^2*tan(f*x)^6*tan(e)^9 - 2 6460*b^2*f*x*tan(f*x)^6*tan(e)^6 + 63*b^2*tan(f*x)^9*tan(e)^4 + 945*b^2*ta n(f*x)^8*tan(e)^5 + 11340*b^2*tan(f*x)^7*tan(e)^6 + 11340*b^2*tan(f*x)^6*t an(e)^7 + 945*b^2*tan(f*x)^5*tan(e)^8 + 63*b^2*tan(f*x)^4*tan(e)^9 + 39690 *b^2*f*x*tan(f*x)^5*tan(e)^5 - 45*b^2*tan(f*x)^9*tan(e)^2 - 567*b^2*tan(f* x)^8*tan(e)^3 - 3780*b^2*tan(f*x)^7*tan(e)^4 - 26460*b^2*tan(f*x)^6*tan(e) ^5 - 26460*b^2*tan(f*x)^5*tan(e)^6 - 3780*b^2*tan(f*x)^4*tan(e)^7 - 567*b^ 2*tan(f*x)^3*tan(e)^8 - 45*b^2*tan(f*x)^2*tan(e)^9 - 39690*b^2*f*x*tan(f*x )^4*tan(e)^4 + 35*b^2*tan(f*x)^9 + 405*b^2*tan(f*x)^8*tan(e) + 2268*b^2*ta n(f*x)^7*tan(e)^2 + 8820*b^2*tan(f*x)^6*tan(e)^3 + 39690*b^2*tan(f*x)^5*ta n(e)^4 + 39690*b^2*tan(f*x)^4*tan(e)^5 + 8820*b^2*tan(f*x)^3*tan(e)^6 + 22 68*b^2*tan(f*x)^2*tan(e)^7 + 405*b^2*tan(f*x)*tan(e)^8 + 35*b^2*tan(e)^9 + 26460*b^2*f*x*tan(f*x)^3*tan(e)^3 - 45*b^2*tan(f*x)^7 - 567*b^2*tan(f*x)^ 6*tan(e) - 3780*b^2*tan(f*x)^5*tan(e)^2 - 26460*b^2*tan(f*x)^4*tan(e)^3 - 26460*b^2*tan(f*x)^3*tan(e)^4 - 3780*b^2*tan(f*x)^2*tan(e)^5 - 567*b^2*tan (f*x)*tan(e)^6 - 45*b^2*tan(e)^7 - 11340*b^2*f*x*tan(f*x)^2*tan(e)^2 + 63* b^2*tan(f*x)^5 + 945*b^2*tan(f*x)^4*tan(e) + 11340*b^2*tan(f*x)^3*tan(e...
Timed out. \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{5/2} \,d x \]