Integrand size = 16, antiderivative size = 88 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\tan (c+d x)}{5 d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {4 \tan (c+d x)}{15 a d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {8 \tan (c+d x)}{15 a^2 d \sqrt {a \sec ^2(c+d x)}} \] Output:
1/5*tan(d*x+c)/d/(a*sec(d*x+c)^2)^(5/2)+4/15*tan(d*x+c)/a/d/(a*sec(d*x+c)^ 2)^(3/2)+8/15*tan(d*x+c)/a^2/d/(a*sec(d*x+c)^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\left (15-10 \sin ^2(c+d x)+3 \sin ^4(c+d x)\right ) \tan (c+d x)}{15 a^2 d \sqrt {a \sec ^2(c+d x)}} \] Input:
Integrate[(a + a*Tan[c + d*x]^2)^(-5/2),x]
Output:
((15 - 10*Sin[c + d*x]^2 + 3*Sin[c + d*x]^4)*Tan[c + d*x])/(15*a^2*d*Sqrt[ a*Sec[c + d*x]^2])
Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4140, 3042, 4610, 209, 209, 208}
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.
\(\displaystyle \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \tan (c+d x)^2+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4140 |
\(\displaystyle \int \frac {1}{\left (a \sec ^2(c+d x)\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \sec (c+d x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle \frac {a \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{7/2}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {a \left (\frac {4 \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{5/2}}d\tan (c+d x)}{5 a}+\frac {\tan (c+d x)}{5 a \left (a \tan ^2(c+d x)+a\right )^{5/2}}\right )}{d}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {a \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{3/2}}d\tan (c+d x)}{3 a}+\frac {\tan (c+d x)}{3 a \left (a \tan ^2(c+d x)+a\right )^{3/2}}\right )}{5 a}+\frac {\tan (c+d x)}{5 a \left (a \tan ^2(c+d x)+a\right )^{5/2}}\right )}{d}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {a \left (\frac {4 \left (\frac {2 \tan (c+d x)}{3 a^2 \sqrt {a \tan ^2(c+d x)+a}}+\frac {\tan (c+d x)}{3 a \left (a \tan ^2(c+d x)+a\right )^{3/2}}\right )}{5 a}+\frac {\tan (c+d x)}{5 a \left (a \tan ^2(c+d x)+a\right )^{5/2}}\right )}{d}\) |
Input:
Int[(a + a*Tan[c + d*x]^2)^(-5/2),x]
Output:
(a*(Tan[c + d*x]/(5*a*(a + a*Tan[c + d*x]^2)^(5/2)) + (4*(Tan[c + d*x]/(3* a*(a + a*Tan[c + d*x]^2)^(3/2)) + (2*Tan[c + d*x])/(3*a^2*Sqrt[a + a*Tan[c + d*x]^2])))/(5*a)))/d
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*sec[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a, b]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFac tors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] && !IntegerQ[p]
Time = 0.82 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\tan \left (d x +c \right )}{5 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}}{a}\right )}{d}\) | \(88\) |
default | \(\frac {a \left (\frac {\tan \left (d x +c \right )}{5 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}}{a}\right )}{d}\) | \(88\) |
risch | \(-\frac {i {\mathrm e}^{6 i \left (d x +c \right )}}{160 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}-\frac {5 i {\mathrm e}^{2 i \left (d x +c \right )}}{16 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}+\frac {5 i}{16 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {5 i {\mathrm e}^{-2 i \left (d x +c \right )}}{96 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}-\frac {11 i \cos \left (4 d x +4 c \right )}{240 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {7 \sin \left (4 d x +4 c \right )}{120 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\) | \(334\) |
Input:
int(1/(a+a*tan(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/d*a*(1/5/a*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(5/2)+4/5/a*(1/3/a*tan(d*x+c)/( a+a*tan(d*x+c)^2)^(3/2)+2/3/a^2*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {{\left (8 \, \tan \left (d x + c\right )^{5} + 20 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{15 \, {\left (a^{3} d \tan \left (d x + c\right )^{6} + 3 \, a^{3} d \tan \left (d x + c\right )^{4} + 3 \, a^{3} d \tan \left (d x + c\right )^{2} + a^{3} d\right )}} \] Input:
integrate(1/(a+a*tan(d*x+c)^2)^(5/2),x, algorithm="fricas")
Output:
1/15*(8*tan(d*x + c)^5 + 20*tan(d*x + c)^3 + 15*tan(d*x + c))*sqrt(a*tan(d *x + c)^2 + a)/(a^3*d*tan(d*x + c)^6 + 3*a^3*d*tan(d*x + c)^4 + 3*a^3*d*ta n(d*x + c)^2 + a^3*d)
\[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+a*tan(d*x+c)**2)**(5/2),x)
Output:
Integral((a*tan(c + d*x)**2 + a)**(-5/2), x)
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \sin \left (3 \, d x + 3 \, c\right ) + 150 \, \sin \left (d x + c\right )}{240 \, a^{\frac {5}{2}} d} \] Input:
integrate(1/(a+a*tan(d*x+c)^2)^(5/2),x, algorithm="maxima")
Output:
1/240*(3*sin(5*d*x + 5*c) + 25*sin(3*d*x + 3*c) + 150*sin(d*x + c))/(a^(5/ 2)*d)
Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (76) = 152\).
Time = 2.55 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.82 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=-\frac {8 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{5} + 20 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{3} + {\left ({\left (20 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{5} + 50 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{3} + {\left (50 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{2} + {\left (\frac {15 \, \tan \left (c\right )^{5}}{\mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right )} + {\left (15 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{4} + 20 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )^{2} + 8 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right )\right )} \tan \left (d x\right )\right )} \tan \left (d x\right ) + 20 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right )\right )} \tan \left (d x\right )\right )} \tan \left (d x\right ) + \frac {15}{\mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right )}\right )} \tan \left (d x\right ) + 15 \, \mathrm {sgn}\left (\tan \left (d x\right ) \tan \left (c\right ) - 1\right ) \tan \left (c\right )}{15 \, {\left (a \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + a \tan \left (d x\right )^{2} + a \tan \left (c\right )^{2} + a\right )}^{\frac {5}{2}} d} \] Input:
integrate(1/(a+a*tan(d*x+c)^2)^(5/2),x, algorithm="giac")
Output:
-1/15*(8*sgn(tan(d*x)*tan(c) - 1)*tan(c)^5 + 20*sgn(tan(d*x)*tan(c) - 1)*t an(c)^3 + ((20*sgn(tan(d*x)*tan(c) - 1)*tan(c)^5 + 50*sgn(tan(d*x)*tan(c) - 1)*tan(c)^3 + (50*sgn(tan(d*x)*tan(c) - 1)*tan(c)^2 + (15*tan(c)^5/sgn(t an(d*x)*tan(c) - 1) + (15*sgn(tan(d*x)*tan(c) - 1)*tan(c)^4 + 20*sgn(tan(d *x)*tan(c) - 1)*tan(c)^2 + 8*sgn(tan(d*x)*tan(c) - 1))*tan(d*x))*tan(d*x) + 20*sgn(tan(d*x)*tan(c) - 1))*tan(d*x))*tan(d*x) + 15/sgn(tan(d*x)*tan(c) - 1))*tan(d*x) + 15*sgn(tan(d*x)*tan(c) - 1)*tan(c))/((a*tan(d*x)^2*tan(c )^2 + a*tan(d*x)^2 + a*tan(c)^2 + a)^(5/2)*d)
Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,{\mathrm {tan}\left (c+d\,x\right )}^4+20\,{\mathrm {tan}\left (c+d\,x\right )}^2+15\right )}{15\,d\,{\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}^{5/2}} \] Input:
int(1/(a + a*tan(c + d*x)^2)^(5/2),x)
Output:
(tan(c + d*x)*(20*tan(c + d*x)^2 + 8*tan(c + d*x)^4 + 15))/(15*d*(a + a*ta n(c + d*x)^2)^(5/2))
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\sqrt {a}\, \sqrt {\tan \left (d x +c \right )^{2}+1}\, \tan \left (d x +c \right ) \left (8 \tan \left (d x +c \right )^{4}+20 \tan \left (d x +c \right )^{2}+15\right )}{15 a^{3} d \left (\tan \left (d x +c \right )^{6}+3 \tan \left (d x +c \right )^{4}+3 \tan \left (d x +c \right )^{2}+1\right )} \] Input:
int(1/(a+a*tan(d*x+c)^2)^(5/2),x)
Output:
(sqrt(a)*sqrt(tan(c + d*x)**2 + 1)*tan(c + d*x)*(8*tan(c + d*x)**4 + 20*ta n(c + d*x)**2 + 15))/(15*a**3*d*(tan(c + d*x)**6 + 3*tan(c + d*x)**4 + 3*t an(c + d*x)**2 + 1))