Integrand size = 16, antiderivative size = 134 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac {b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(5 a-2 b) b \tan (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f-1/3* (5*a-2*b)*b*tan(f*x+e)/a^2/(a-b)^2/f/(a+b*tan(f*x+e)^2)^(1/2)-1/3*b*tan(f* x+e)/a/(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 = 7.72 (sec) , antiderivative size = 1331, normalized size of antiderivative = 9.93 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \]
(Cos[e + f*x]*Sin[e + f*x]*(1575*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]] - (3150*(a - b)*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^2)/a + (1575*(a - b)^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^4 )/a^2 + (2100*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/a - (4200*(a - b)*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^2 *Tan[e + f*x]^2)/a^2 + (2100*(a - b)^2*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x] ^2)/a]]*Sin[e + f*x]^4*Tan[e + f*x]^2)/a^3 + (840*b^2*ArcSin[Sqrt[((a - b) *Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4)/a^2 - (1680*(a - b)*b^2*ArcSin[Sqrt[( (a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^2*Tan[e + f*x]^4)/a^3 + (840*(a - b)^2*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^4*Tan[e + f*x]^4)/a^4 + 2100*(((a - b)*Sin[e + f*x]^2)/a)^(3/2)*Sqrt[(Cos[e + f*x]^2 *(a + b*Tan[e + f*x]^2))/a] + 96*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin [e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a] + 24*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Sqrt[(Cos[e + f* x]^2*(a + b*Tan[e + f*x]^2))/a] + (2800*b*(((a - b)*Sin[e + f*x]^2)/a)^(3/ 2)*Tan[e + f*x]^2*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a])/a + (16 8*b*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*(((a - b)*Sin [e + f*x]^2)/a)^(7/2)*Tan[e + f*x]^2*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f *x]^2))/a])/a + (48*b*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*S...
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4144, 316, 402, 27, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-2 b \tan ^2(e+f x)+3 a-2 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{3 a (a-b)}-\frac {b \tan (e+f x)}{3 a (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a (a-b)}-\frac {b (5 a-2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{3 a (a-b)}-\frac {b \tan (e+f x)}{3 a (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 a \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a-b}-\frac {b (5 a-2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{3 a (a-b)}-\frac {b \tan (e+f x)}{3 a (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {3 a \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{a-b}-\frac {b (5 a-2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{3 a (a-b)}-\frac {b \tan (e+f x)}{3 a (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {3 a \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2}}-\frac {b (5 a-2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{3 a (a-b)}-\frac {b \tan (e+f x)}{3 a (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
(-1/3*(b*Tan[e + f*x])/(a*(a - b)*(a + b*Tan[e + f*x]^2)^(3/2)) + ((3*a*Ar cTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(a - b)^(3/2) - ((5*a - 2*b)*b*Tan[e + f*x])/(a*(a - b)*Sqrt[a + b*Tan[e + f*x]^2]))/(3 *a*(a - b)))/f
3.2.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[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 = 0.06 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.22
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\tan \left (f x +e \right )}{3 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{a -b}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{3} b^{2}}-\frac {b \tan \left (f x +e \right )}{\left (a -b \right )^{2} a \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(163\) |
| default | \(\frac {-\frac {b \left (\frac {\tan \left (f x +e \right )}{3 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{a -b}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{3} b^{2}}-\frac {b \tan \left (f x +e \right )}{\left (a -b \right )^{2} a \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(163\) |
1/f*(-b/(a-b)*(1/3*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(3/2)+2/3/a^2*tan(f*x+e )/(a+b*tan(f*x+e)^2)^(1/2))+1/(a-b)^3*(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)^2*tan(f* x+e)/a/(a+b*tan(f*x+e)^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (120) = 240\).
Time = 0.34 (sec) , antiderivative size = 561, normalized size of antiderivative = 4.19 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (a^{2} b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{3} b \tan \left (f x + e\right )^{2} + a^{4}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left ({\left (5 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}, \frac {3 \, {\left (a^{2} b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{3} b \tan \left (f x + e\right )^{2} + a^{4}\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - {\left ({\left (5 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{3 \, {\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}\right ] \]
[-1/6*(3*(a^2*b^2*tan(f*x + e)^4 + 2*a^3*b*tan(f*x + e)^2 + a^4)*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*((5*a^2*b^2 - 7*a*b^3 + 2 *b^4)*tan(f*x + e)^3 + 3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*tan(f*x + e))*sqrt( b*tan(f*x + e)^2 + a))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*f*tan( f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*tan(f*x + e)^2 + (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f), 1/3*(3*(a^2*b^2*tan(f*x + e)^4 + 2*a^3*b*tan(f*x + e)^2 + a^4)*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - ((5*a^2*b^2 - 7*a*b^3 + 2*b^4)*tan(f*x + e)^3 + 3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e )^2 + a))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*f*tan(f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*tan(f*x + e)^2 + (a^7 - 3*a^ 6*b + 3*a^5*b^2 - a^4*b^3)*f)]
\[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]