Integrand size = 23, antiderivative size = 99 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {1}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {1}{(a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
-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-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 = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right )}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[e + f*x]^2)/(a - b)]/(3*(a - b )*f*(a + b*Tan[e + f*x]^2)^(3/2))
Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4153, 353, 61, 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.
\(\displaystyle \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)}{\left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan (e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan ^2(e+f x)}{a-b}+\frac {2}{3 (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan ^2(e+f x)}{a-b}+\frac {2}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{a-b}+\frac {2}{3 (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\frac {2 \int \frac {1}{\frac {\tan ^4(e+f x)}{b}-\frac {a}{b}+1}d\sqrt {b \tan ^2(e+f x)+a}}{b (a-b)}+\frac {2}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{a-b}+\frac {2}{3 (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {2}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}}{a-b}+\frac {2}{3 (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{2 f}\) |
(2/(3*(a - 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)
3.4.48.3.1 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_)^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_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
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 = 89, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\frac {1}{3 \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}+\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(89\) |
default | \(\frac {\frac {1}{3 \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}+\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(89\) |
1/f*(1/3/(a-b)/(a+b*tan(f*x+e)^2)^(3/2)+1/(a-b)^2/(-a+b)^(1/2)*arctan((a+b *tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2))+1/(a-b)^2/(a+b*tan(f*x+e)^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (87) = 174\).
Time = 0.34 (sec) , antiderivative size = 544, normalized size of antiderivative = 5.49 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (3 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, a^{2} - 5 \, a b + b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}, \frac {3 \, {\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (3 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, a^{2} - 5 \, a b + b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}\right ] \]
[1/12*(3*(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)*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*(3*(a*b - b^2)*tan(f*x + e)^2 + 4*a^2 - 5*a*b + b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^2 - 3*a^2* b^3 + 3*a*b^4 - b^5)*f*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f), 1/6*( 3*(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)*sqrt(-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* (3*(a*b - b^2)*tan(f*x + e)^2 + 4*a^2 - 5*a*b + b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3 *b^2 - a^2*b^3)*f)]
Time = 12.77 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.39 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b}{6 f \left (a - b\right ) \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}} + \frac {b}{2 f \left (a - b\right )^{2} \sqrt {a + b \tan ^{2}{\left (e + f x \right )}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}{\sqrt {- a + b}} \right )}}{2 f \sqrt {- a + b} \left (a - b\right )^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \tilde {\infty } \tan ^{2}{\left (e + f x \right )} & \text {for}\: a^{\frac {5}{2}} = 0 \vee f = 0 \\\frac {\log {\left (2 a^{\frac {5}{2}} f \tan ^{2}{\left (e + f x \right )} + 2 a^{\frac {5}{2}} f \right )}}{2 a^{\frac {5}{2}} f} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]
Piecewise((2*(b/(6*f*(a - b)*(a + b*tan(e + f*x)**2)**(3/2)) + b/(2*f*(a - b)**2*sqrt(a + b*tan(e + f*x)**2)) + b*atan(sqrt(a + b*tan(e + f*x)**2)/s qrt(-a + b))/(2*f*sqrt(-a + b)*(a - b)**2))/b, Ne(b, 0)), (Piecewise((zoo* tan(e + f*x)**2, Eq(f, 0) | Eq(a**(5/2), 0)), (log(2*a**(5/2)*f*tan(e + f* x)**2 + 2*a**(5/2)*f)/(2*a**(5/2)*f), True)), True))
\[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Time = 16.00 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.32 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}{{\left (a-b\right )}^2}+\frac {1}{3\,\left (a-b\right )}}{f\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {a^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}+b^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-a\,b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,2{}\mathrm {i}}{{\left (a-b\right )}^{5/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{5/2}} \]