Integrand size = 25, antiderivative size = 103 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}-\frac {a}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {1}{(a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/f-1/(a-b)^2/f/(a +b*tan(f*x+e)^2)^(1/2)-1/3*a/(a-b)/b/f/(a+b*tan(f*x+e)^2)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {a (-a+b)-3 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )}{3 (a-b)^2 b f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
(a*(-a + b) - 3*b*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[e + f*x]^2)/( a - b)]*(a + b*Tan[e + f*x]^2))/(3*(a - b)^2*b*f*(a + b*Tan[e + f*x]^2)^(3 /2))
Time = 0.31 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4153, 354, 87, 61, 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.
| \(\displaystyle \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^3}{\left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tan ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {-\frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan ^2(e+f x)}{a-b}-\frac {2 a}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan ^2(e+f x)}{a-b}+\frac {2}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{a-b}-\frac {2 a}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {-\frac {\frac {2 \int \frac {1}{\frac {\tan ^4(e+f x)}{b}-\frac {a}{b}+1}d\sqrt {b \tan ^2(e+f x)+a}}{b (a-b)}+\frac {2}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{a-b}-\frac {2 a}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {\frac {2}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}}{a-b}-\frac {2 a}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
((-2*a)/(3*(a - b)*b*(a + b*Tan[e + f*x]^2)^(3/2)) - ((-2*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/(a - b)^(3/2) + 2/((a - b)*Sqrt[a + b*Tan [e + f*x]^2]))/(a - b))/(2*f)
3.4.47.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.07
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{3 b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{3 \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}}{f}\) | \(110\) |
| default | \(\frac {-\frac {1}{3 b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{3 \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}}{f}\) | \(110\) |
1/f*(-1/3/b/(a+b*tan(f*x+e)^2)^(3/2)-1/3/(a-b)/(a+b*tan(f*x+e)^2)^(3/2)-1/ (a-b)^2/(a+b*tan(f*x+e)^2)^(1/2)-1/(a-b)^2/(-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. 252 vs. \(2 (91) = 182\).
Time = 0.34 (sec) , antiderivative size = 572, normalized size of antiderivative = 5.55 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{3} \tan \left (f x + e\right )^{4} + 2 \, a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (a^{3} + a^{2} b - 2 \, a b^{2} + 3 \, {\left (a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{3} b^{3} - 3 \, a^{2} b^{4} + 3 \, a b^{5} - b^{6}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f\right )}}, -\frac {3 \, {\left (b^{3} \tan \left (f x + e\right )^{4} + 2 \, a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (a^{3} + a^{2} b - 2 \, a b^{2} + 3 \, {\left (a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{3} b^{3} - 3 \, a^{2} b^{4} + 3 \, a b^{5} - b^{6}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f\right )}}\right ] \]
[1/12*(3*(b^3*tan(f*x + e)^4 + 2*a*b^2*tan(f*x + e)^2 + a^2*b)*sqrt(a - 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*(a^3 + a^2*b - 2*a*b^ 2 + 3*(a*b^2 - b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^3 - 3*a^2*b^4 + 3*a*b^5 - b^6)*f*tan(f*x + e)^4 + 2*(a^4*b^2 - 3*a^3*b^3 + 3 *a^2*b^4 - a*b^5)*f*tan(f*x + e)^2 + (a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2* b^4)*f), -1/6*(3*(b^3*tan(f*x + e)^4 + 2*a*b^2*tan(f*x + e)^2 + a^2*b)*sqr t(-a + b)*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e) ^2 + 2*a - b)) + 2*(a^3 + a^2*b - 2*a*b^2 + 3*(a*b^2 - b^3)*tan(f*x + e)^2 )*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^3 - 3*a^2*b^4 + 3*a*b^5 - b^6)*f*tan (f*x + e)^4 + 2*(a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*f*tan(f*x + e)^2 + (a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*f)]
\[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Time = 15.61 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.34 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {a}{3\,\left (a-b\right )}+\frac {b\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}{{\left (a-b\right )}^2}}{b\,f\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}-\frac {\mathrm {atan}\left (\frac {a^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}+b^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-a\,b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,2{}\mathrm {i}}{{\left (a-b\right )}^{5/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{5/2}} \]