Integrand size = 14, antiderivative size = 98 \[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=-\frac {b^2 \cot (e+f x) \log (\cos (e+f x)) \sqrt {b \tan ^2(e+f x)}}{f}-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^2(e+f x)}}{4 f} \]
-b^2*cot(f*x+e)*ln(cos(f*x+e))*(b*tan(f*x+e)^2)^(1/2)/f-1/2*b^2*(b*tan(f*x +e)^2)^(1/2)*tan(f*x+e)/f+1/4*b^2*(b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/f
Time = 0.40 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57 \[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=-\frac {\cot (e+f x) \left (-1+2 \cot ^2(e+f x)+4 \cot ^4(e+f x) \log (\cos (e+f x))\right ) \left (b \tan ^2(e+f x)\right )^{5/2}}{4 f} \]
-1/4*(Cot[e + f*x]*(-1 + 2*Cot[e + f*x]^2 + 4*Cot[e + f*x]^4*Log[Cos[e + f *x]])*(b*Tan[e + f*x]^2)^(5/2))/f
Time = 0.38 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 3042, 3956}
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.
\(\displaystyle \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan (e+f x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \int \tan ^5(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \int \tan (e+f x)^5dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \left (\frac {\tan ^4(e+f x)}{4 f}-\int \tan ^3(e+f x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \left (\frac {\tan ^4(e+f x)}{4 f}-\int \tan (e+f x)^3dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \left (\int \tan (e+f x)dx+\frac {\tan ^4(e+f x)}{4 f}-\frac {\tan ^2(e+f x)}{2 f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \left (\int \tan (e+f x)dx+\frac {\tan ^4(e+f x)}{4 f}-\frac {\tan ^2(e+f x)}{2 f}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \left (\frac {\tan ^4(e+f x)}{4 f}-\frac {\tan ^2(e+f x)}{2 f}-\frac {\log (\cos (e+f x))}{f}\right )\) |
b^2*Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^2]*(-(Log[Cos[e + f*x]]/f) - Tan[e + f*x]^2/(2*f) + Tan[e + f*x]^4/(4*f))
3.1.1.3.1 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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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 = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59
method | result | size |
derivativedivides | \(\frac {\left (b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}} \left (\tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{2}+2 \ln \left (1+\tan \left (f x +e \right )^{2}\right )\right )}{4 f \tan \left (f x +e \right )^{5}}\) | \(58\) |
default | \(\frac {\left (b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}} \left (\tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{2}+2 \ln \left (1+\tan \left (f x +e \right )^{2}\right )\right )}{4 f \tan \left (f x +e \right )^{5}}\) | \(58\) |
risch | \(\frac {b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, x}{{\mathrm e}^{2 i \left (f x +e \right )}-1}-\frac {2 b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, \left (f x +e \right )}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) f}-\frac {4 i b^{2} \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{6 i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} f}-\frac {i b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) f}\) | \(300\) |
1/4/f*(b*tan(f*x+e)^2)^(5/2)*(tan(f*x+e)^4-2*tan(f*x+e)^2+2*ln(1+tan(f*x+e )^2))/tan(f*x+e)^5
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=\frac {{\left (b^{2} \tan \left (f x + e\right )^{4} - 2 \, b^{2} \tan \left (f x + e\right )^{2} - 2 \, b^{2} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2}}}{4 \, f \tan \left (f x + e\right )} \]
1/4*(b^2*tan(f*x + e)^4 - 2*b^2*tan(f*x + e)^2 - 2*b^2*log(1/(tan(f*x + e) ^2 + 1)) - 3*b^2)*sqrt(b*tan(f*x + e)^2)/(f*tan(f*x + e))
\[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=\int \left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
Time = 0.34 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.48 \[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=\frac {b^{\frac {5}{2}} \tan \left (f x + e\right )^{4} - 2 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{2} + 2 \, b^{\frac {5}{2}} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{4 \, f} \]
1/4*(b^(5/2)*tan(f*x + e)^4 - 2*b^(5/2)*tan(f*x + e)^2 + 2*b^(5/2)*log(tan (f*x + e)^2 + 1))/f
Leaf count of result is larger than twice the leaf count of optimal. 646 vs. \(2 (88) = 176\).
Time = 1.00 (sec) , antiderivative size = 646, normalized size of antiderivative = 6.59 \[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=\text {Too large to display} \]
-1/4*(2*b^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^ 2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*sgn(tan(f*x + e))*tan(f*x)^4*tan( e)^4 + 3*b^2*sgn(tan(f*x + e))*tan(f*x)^4*tan(e)^4 - 8*b^2*log(4*(tan(f*x) ^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + t an(e)^2 + 1))*sgn(tan(f*x + e))*tan(f*x)^3*tan(e)^3 + 2*b^2*sgn(tan(f*x + e))*tan(f*x)^4*tan(e)^2 - 8*b^2*sgn(tan(f*x + e))*tan(f*x)^3*tan(e)^3 + 2* b^2*sgn(tan(f*x + e))*tan(f*x)^2*tan(e)^4 + 12*b^2*log(4*(tan(f*x)^2*tan(e )^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*sgn(tan(f*x + e))*tan(f*x)^2*tan(e)^2 - b^2*sgn(tan(f*x + e))*tan(f* x)^4 - 8*b^2*sgn(tan(f*x + e))*tan(f*x)^3*tan(e) + 4*b^2*sgn(tan(f*x + e)) *tan(f*x)^2*tan(e)^2 - 8*b^2*sgn(tan(f*x + e))*tan(f*x)*tan(e)^3 - b^2*sgn (tan(f*x + e))*tan(e)^4 - 8*b^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*ta n(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*sgn(tan(f*x + e))*tan(f*x)*tan(e) + 2*b^2*sgn(tan(f*x + e))*tan(f*x)^2 - 8*b^2*sgn(tan( f*x + e))*tan(f*x)*tan(e) + 2*b^2*sgn(tan(f*x + e))*tan(e)^2 + 2*b^2*log(4 *(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan( f*x)^2 + tan(e)^2 + 1))*sgn(tan(f*x + e)) + 3*b^2*sgn(tan(f*x + e)))*sqrt( b)/(f*tan(f*x)^4*tan(e)^4 - 4*f*tan(f*x)^3*tan(e)^3 + 6*f*tan(f*x)^2*tan(e )^2 - 4*f*tan(f*x)*tan(e) + f)
Timed out. \[ \int \left (b \tan ^2(e+f x)\right )^{5/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{5/2} \,d x \]