Integrand size = 25, antiderivative size = 81 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}+\frac {\tan (e+f x)}{(a-b) f \sqrt {a+b \tan ^2(e+f x)}} \]
-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f+tan (f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)
Time = 3.48 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.90 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\tan (e+f x) \left (\text {arctanh}\left (\frac {\sqrt {\frac {(-a+b) \tan ^2(e+f x)}{a}}}{\sqrt {1+\frac {b \tan ^2(e+f x)}{a}}}\right ) \left (b+a \cot ^2(e+f x)\right ) \sqrt {\frac {(-a+b) \tan ^2(e+f x)}{a}}+(a-b) \sqrt {1+\frac {b \tan ^2(e+f x)}{a}}\right )}{(a-b)^2 f \sqrt {a+b \tan ^2(e+f x)} \sqrt {1+\frac {b \tan ^2(e+f x)}{a}}} \]
(Tan[e + f*x]*(ArcTanh[Sqrt[((-a + b)*Tan[e + f*x]^2)/a]/Sqrt[1 + (b*Tan[e + f*x]^2)/a]]*(b + a*Cot[e + f*x]^2)*Sqrt[((-a + b)*Tan[e + f*x]^2)/a] + (a - b)*Sqrt[1 + (b*Tan[e + f*x]^2)/a]))/((a - b)^2*f*Sqrt[a + b*Tan[e + f *x]^2]*Sqrt[1 + (b*Tan[e + f*x]^2)/a])
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4153, 373, 291, 216}
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.
\(\displaystyle \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^2}{\left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 373 |
\(\displaystyle \frac {\frac {\tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a-b}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{a-b}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2}}}{f}\) |
(-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(a - b)^( 3/2)) + Tan[e + f*x]/((a - b)*Sqrt[a + b*Tan[e + f*x]^2]))/f
3.4.41.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
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 = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )}{a \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {b \tan \left (f x +e \right )}{\left (a -b \right ) a \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}}{f}\) | \(126\) |
default | \(\frac {\frac {\tan \left (f x +e \right )}{a \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {b \tan \left (f x +e \right )}{\left (a -b \right ) a \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}}{f}\) | \(126\) |
1/f*(tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(1/2)+b/(a-b)*tan(f*x+e)/a/(a+b*tan(f *x+e)^2)^(1/2)-1/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b) )^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)))
Time = 0.29 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.52 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (b \tan \left (f x + e\right )^{2} + a\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )} \tan \left (f x + e\right )}{2 \, {\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} f\right )}}, -\frac {{\left (b \tan \left (f x + e\right )^{2} + a\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )} \tan \left (f x + e\right )}{{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} f}\right ] \]
[1/2*((b*tan(f*x + e)^2 + a)*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e) ^2 + 1)) + 2*sqrt(b*tan(f*x + e)^2 + a)*(a - b)*tan(f*x + e))/((a^2*b - 2* a*b^2 + b^3)*f*tan(f*x + e)^2 + (a^3 - 2*a^2*b + a*b^2)*f), -((b*tan(f*x + e)^2 + a)*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan (f*x + e))) - sqrt(b*tan(f*x + e)^2 + a)*(a - b)*tan(f*x + e))/((a^2*b - 2 *a*b^2 + b^3)*f*tan(f*x + e)^2 + (a^3 - 2*a^2*b + a*b^2)*f)]
\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Timed out. \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]