Integrand size = 14, antiderivative size = 98 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=-\frac {b^2 \cot (c+d x) \log (\cos (c+d x)) \sqrt {b \tan ^2(c+d x)}}{d}-\frac {b^2 \tan (c+d x) \sqrt {b \tan ^2(c+d x)}}{2 d}+\frac {b^2 \tan ^3(c+d x) \sqrt {b \tan ^2(c+d x)}}{4 d} \] Output:
-b^2*cot(d*x+c)*ln(cos(d*x+c))*(b*tan(d*x+c)^2)^(1/2)/d-1/2*b^2*tan(d*x+c) *(b*tan(d*x+c)^2)^(1/2)/d+1/4*b^2*tan(d*x+c)^3*(b*tan(d*x+c)^2)^(1/2)/d
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {b^2 \cot (c+d x) \left (-4 \log (\cos (c+d x))-4 \sec ^2(c+d x)+\sec ^4(c+d x)\right ) \sqrt {b \tan ^2(c+d x)}}{4 d} \] Input:
Integrate[(b*Tan[c + d*x]^2)^(5/2),x]
Output:
(b^2*Cot[c + d*x]*(-4*Log[Cos[c + d*x]] - 4*Sec[c + d*x]^2 + Sec[c + d*x]^ 4)*Sqrt[b*Tan[c + d*x]^2])/(4*d)
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.
\(\displaystyle \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan (c+d x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \int \tan ^5(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \int \tan (c+d x)^5dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan ^3(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan (c+d x)^3dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\) |
Input:
Int[(b*Tan[c + d*x]^2)^(5/2),x]
Output:
b^2*Cot[c + d*x]*Sqrt[b*Tan[c + d*x]^2]*(-(Log[Cos[c + d*x]]/d) - Tan[c + d*x]^2/(2*d) + Tan[c + d*x]^4/(4*d))
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 = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59
method | result | size |
derivativedivides | \(\frac {\left (b \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}} \left (\tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{2}+2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )\right )}{4 d \tan \left (d x +c \right )^{5}}\) | \(58\) |
default | \(\frac {\left (b \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}} \left (\tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{2}+2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )\right )}{4 d \tan \left (d x +c \right )^{5}}\) | \(58\) |
risch | \(\frac {b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, x}{{\mathrm e}^{2 i \left (d x +c \right )}-1}-\frac {2 b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left (d x +c \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}-\frac {4 i b^{2} \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{6 i \left (d x +c \right )}+{\mathrm e}^{4 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}-\frac {i b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}\) | \(300\) |
Input:
int((b*tan(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/4/d*(b*tan(d*x+c)^2)^(5/2)*(tan(d*x+c)^4-2*tan(d*x+c)^2+2*ln(1+tan(d*x+c )^2))/tan(d*x+c)^5
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {{\left (b^{2} \tan \left (d x + c\right )^{4} - 2 \, b^{2} \tan \left (d x + c\right )^{2} - 2 \, b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt {b \tan \left (d x + c\right )^{2}}}{4 \, d \tan \left (d x + c\right )} \] Input:
integrate((b*tan(d*x+c)^2)^(5/2),x, algorithm="fricas")
Output:
1/4*(b^2*tan(d*x + c)^4 - 2*b^2*tan(d*x + c)^2 - 2*b^2*log(1/(tan(d*x + c) ^2 + 1)) - 3*b^2)*sqrt(b*tan(d*x + c)^2)/(d*tan(d*x + c))
\[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\int \left (b \tan ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((b*tan(d*x+c)**2)**(5/2),x)
Output:
Integral((b*tan(c + d*x)**2)**(5/2), x)
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.48 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {b^{\frac {5}{2}} \tan \left (d x + c\right )^{4} - 2 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{2} + 2 \, b^{\frac {5}{2}} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{4 \, d} \] Input:
integrate((b*tan(d*x+c)^2)^(5/2),x, algorithm="maxima")
Output:
1/4*(b^(5/2)*tan(d*x + c)^4 - 2*b^(5/2)*tan(d*x + c)^2 + 2*b^(5/2)*log(tan (d*x + c)^2 + 1))/d
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.56 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {1}{4} \, b^{\frac {5}{2}} {\left (\frac {2 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{d} + \frac {d \tan \left (d x + c\right )^{4} - 2 \, d \tan \left (d x + c\right )^{2}}{d^{2}}\right )} \mathrm {sgn}\left (\tan \left (d x + c\right )\right ) \] Input:
integrate((b*tan(d*x+c)^2)^(5/2),x, algorithm="giac")
Output:
1/4*b^(5/2)*(2*log(tan(d*x + c)^2 + 1)/d + (d*tan(d*x + c)^4 - 2*d*tan(d*x + c)^2)/d^2)*sgn(tan(d*x + c))
Timed out. \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}^{5/2} \,d x \] Input:
int((b*tan(c + d*x)^2)^(5/2),x)
Output:
int((b*tan(c + d*x)^2)^(5/2), x)
Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.43 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {\sqrt {b}\, b^{2} \left (2 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+\tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{2}\right )}{4 d} \] Input:
int((b*tan(d*x+c)^2)^(5/2),x)
Output:
(sqrt(b)*b**2*(2*log(tan(c + d*x)**2 + 1) + tan(c + d*x)**4 - 2*tan(c + d* x)**2))/(4*d)