Integrand size = 16, antiderivative size = 85 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \tan (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}} \] Output:
arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-b*ta n(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 4.61 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.73 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\cos (e+f x) \sin (e+f x) \left (\frac {4 (a-b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}-\frac {15 \left (2 b+3 a \cot ^2(e+f x)\right ) \left (-a \sec ^2(e+f x) \sqrt {\frac {(a-b) \left (b+a \cot ^2(e+f x)\right ) \sin ^4(e+f x)}{a^2}}+\arcsin \left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a+b \tan ^2(e+f x)\right )\right )}{a (a-b) \sqrt {\frac {(a-b) \left (b+a \cot ^2(e+f x)\right ) \sin ^4(e+f x)}{a^2}}}\right )}{15 a f \sqrt {a+b \tan ^2(e+f x)}} \] Input:
Integrate[(a + b*Tan[e + f*x]^2)^(-3/2),x]
Output:
(Cos[e + f*x]*Sin[e + f*x]*((4*(a - b)*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2 - (15*(2* b + 3*a*Cot[e + f*x]^2)*(-(a*Sec[e + f*x]^2*Sqrt[((a - b)*(b + a*Cot[e + f *x]^2)*Sin[e + f*x]^4)/a^2]) + ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*(a + b*Tan[e + f*x]^2)))/(a*(a - b)*Sqrt[((a - b)*(b + a*Cot[e + f*x]^2)*Sin [e + f*x]^4)/a^2])))/(15*a*f*Sqrt[a + b*Tan[e + f*x]^2])
Time = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3042, 4144, 296, 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.
\(\displaystyle \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4144 |
\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 296 |
\(\displaystyle \frac {\frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a-b}-\frac {b \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{a-b}-\frac {b \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2}}-\frac {b \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
Input:
Int[(a + b*Tan[e + f*x]^2)^(-3/2),x]
Output:
(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(a - b)^(3/ 2) - (b*Tan[e + f*x])/(a*(a - b)*Sqrt[a + b*Tan[e + f*x]^2]))/f
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Int[((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[(a + b* (ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 0.64 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.20
method | result | size |
derivativedivides | \(\frac {\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}-\frac {b \tan \left (f x +e \right )}{\left (a -b \right ) a \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(102\) |
default | \(\frac {\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}-\frac {b \tan \left (f x +e \right )}{\left (a -b \right ) a \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(102\) |
Input:
int(1/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f*(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))-b/(a-b)*tan(f*x+e)/a/(a+b*tan(f*x+e)^2) ^(1/2))
Time = 0.16 (sec) , antiderivative size = 307, normalized size of antiderivative = 3.61 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{2 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, \frac {{\left (a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {a - b} \arctan \left (\frac {\sqrt {a - b} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) - \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f}\right ] \] Input:
integrate(1/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
[1/2*((a*b*tan(f*x + e)^2 + a^2)*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 - b^2)*tan(f*x + e))/((a^ 3*b - 2*a^2*b^2 + a*b^3)*f*tan(f*x + e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f), ((a*b*tan(f*x + e)^2 + a^2)*sqrt(a - b)*arctan(sqrt(a - b)*tan(f*x + e)/sq rt(b*tan(f*x + e)^2 + a)) - sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*tan(f*x + e))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*tan(f*x + e)^2 + (a^4 - 2*a^3*b + a^ 2*b^2)*f)]
\[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(a+b*tan(f*x+e)**2)**(3/2),x)
Output:
Integral((a + b*tan(e + f*x)**2)**(-3/2), x)
Exception generated. \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
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. \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:
int(1/(a + b*tan(e + f*x)^2)^(3/2),x)
Output:
int(1/(a + b*tan(e + f*x)^2)^(3/2), x)
\[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}}{\tan \left (f x +e \right )^{4} b^{2}+2 \tan \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:
int(1/(a+b*tan(f*x+e)^2)^(3/2),x)
Output:
int(sqrt(tan(e + f*x)**2*b + a)/(tan(e + f*x)**4*b**2 + 2*tan(e + f*x)**2* a*b + a**2),x)