Integrand size = 25, antiderivative size = 64 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b f} \] Output:
arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(1/2)/f+(a+b*tan(f*x+e )^2)^(1/2)/b/f
Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b}}{f} \] Input:
Integrate[Tan[e + f*x]^3/Sqrt[a + b*Tan[e + f*x]^2],x]
Output:
(ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]]/Sqrt[a - b] + Sqrt[a + b* Tan[e + f*x]^2]/b)/f
Time = 0.47 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4153, 354, 90, 73, 221}
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.
\(\displaystyle \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^3}{\sqrt {a+b \tan (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {\frac {2 \sqrt {a+b \tan ^2(e+f x)}}{b}-\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 \sqrt {a+b \tan ^2(e+f x)}}{b}-\frac {2 \int \frac {1}{\frac {\tan ^4(e+f x)}{b}-\frac {a}{b}+1}d\sqrt {b \tan ^2(e+f x)+a}}{b}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \sqrt {a+b \tan ^2(e+f x)}}{b}}{2 f}\) |
Input:
Int[Tan[e + f*x]^3/Sqrt[a + b*Tan[e + f*x]^2],x]
Output:
((2*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/Sqrt[a - b] + (2*Sqrt [a + b*Tan[e + f*x]^2])/b)/(2*f)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.82 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\) | \(56\) |
default | \(\frac {\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\) | \(56\) |
Input:
int(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/b*(a+b*tan(f*x+e)^2)^(1/2)-1/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2) ^(1/2)/(-a+b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (56) = 112\).
Time = 0.13 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.28 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {\sqrt {a - b} b \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )}}{4 \, {\left (a b - b^{2}\right )} f}, \frac {\sqrt {-a + b} b \arctan \left (-\frac {{\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{2 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} - a b\right )}}\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )}}{2 \, {\left (a b - b^{2}\right )} f}\right ] \] Input:
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
[1/4*(sqrt(a - b)*b*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x + e)^2 + 4*(b*tan(f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*sq rt(b*tan(f*x + e)^2 + a)*(a - b))/((a*b - b^2)*f), 1/2*(sqrt(-a + b)*b*arc tan(-1/2*(b*tan(f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/((a*b - b^2)*tan(f*x + e)^2 + a^2 - a*b)) + 2*sqrt(b*tan(f*x + e)^2 + a)*(a - b))/((a*b - b^2)*f)]
\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \] Input:
integrate(tan(f*x+e)**3/(a+b*tan(f*x+e)**2)**(1/2),x)
Output:
Integral(tan(e + f*x)**3/sqrt(a + b*tan(e + f*x)**2), x)
\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{3}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \] Input:
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(tan(f*x + e)^3/sqrt(b*tan(f*x + e)^2 + a), x)
Timed out. \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \] Input:
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
Timed out
Time = 8.72 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )}{f\,\sqrt {a-b}}+\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{b\,f} \] Input:
int(tan(e + f*x)^3/(a + b*tan(e + f*x)^2)^(1/2),x)
Output:
atanh((a + b*tan(e + f*x)^2)^(1/2)/(a - b)^(1/2))/(f*(a - b)^(1/2)) + (a + b*tan(e + f*x)^2)^(1/2)/(b*f)
\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b +a}d x \] Input:
int(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x)
Output:
int((sqrt(tan(e + f*x)**2*b + a)*tan(e + f*x)**3)/(tan(e + f*x)**2*b + a), x)