\(\int \frac {1}{(a+a \tan ^2(c+d x))^{5/2}} \, dx\) [285]

Optimal result

Integrand size = 16, antiderivative size = 88 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\tan (c+d x)}{5 d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {4 \tan (c+d x)}{15 a d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {8 \tan (c+d x)}{15 a^2 d \sqrt {a \sec ^2(c+d x)}} \]

[Out]

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*se
c(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3738, 4207, 198, 197} \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {8 \tan (c+d x)}{15 a^2 d \sqrt {a \sec ^2(c+d x)}}+\frac {4 \tan (c+d x)}{15 a d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {\tan (c+d x)}{5 d \left (a \sec ^2(c+d x)\right )^{5/2}} \]

[In]

Int[(a + a*Tan[c + d*x]^2)^(-5/2),x]

[Out]

Tan[c + d*x]/(5*d*(a*Sec[c + d*x]^2)^(5/2)) + (4*Tan[c + d*x])/(15*a*d*(a*Sec[c + d*x]^2)^(3/2)) + (8*Tan[c +
d*x])/(15*a^2*d*Sqrt[a*Sec[c + d*x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 3738

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4207

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[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
]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (a \sec ^2(c+d x)\right )^{5/2}} \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{7/2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\tan (c+d x)}{5 d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {4 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{5 d} \\ & = \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 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{15 a d} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\left (15-10 \sin ^2(c+d x)+3 \sin ^4(c+d x)\right ) \tan (c+d x)}{15 a^2 d \sqrt {a \sec ^2(c+d x)}} \]

[In]

Integrate[(a + a*Tan[c + d*x]^2)^(-5/2),x]

[Out]

((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])

Maple [A] (verified)

Time = 0.09 (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	\(-\frac {i {\mathrm e}^{6 i \left (d x +c \right )}}{160 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}-\frac {5 i {\mathrm e}^{2 i \left (d x +c \right )}}{16 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}+\frac {5 i}{16 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {5 i {\mathrm e}^{-2 i \left (d x +c \right )}}{96 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}-\frac {11 i \cos \left (4 d x +4 c \right )}{240 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {7 \sin \left (4 d x +4 c \right )}{120 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\)	\(334\)

[In]

int(1/(a+a*tan(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {{\left (8 \, \tan \left (d x + c\right )^{5} + 20 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{15 \, {\left (a^{3} d \tan \left (d x + c\right )^{6} + 3 \, a^{3} d \tan \left (d x + c\right )^{4} + 3 \, a^{3} d \tan \left (d x + c\right )^{2} + a^{3} d\right )}} \]

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(5/2),x, algorithm="fricas")

[Out]

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*tan(d*x + c)^2 + a^3*d)

Sympy [F]

\[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(a+a*tan(d*x+c)**2)**(5/2),x)

[Out]

Integral((a*tan(c + d*x)**2 + a)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \sin \left (3 \, d x + 3 \, c\right ) + 150 \, \sin \left (d x + c\right )}{240 \, a^{\frac {5}{2}} d} \]

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(5/2),x, algorithm="maxima")

[Out]

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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1276 vs. \(2 (76) = 152\).

Time = 1.84 (sec) , antiderivative size = 1276, normalized size of antiderivative = 14.50 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(5/2),x, algorithm="giac")

[Out]

-2/15*(15*tan(1/2*d*x)^9*tan(1/2*c)^10 - 75*tan(1/2*d*x)^9*tan(1/2*c)^8 - 150*tan(1/2*d*x)^8*tan(1/2*c)^9 + 20
*tan(1/2*d*x)^7*tan(1/2*c)^10 + 150*tan(1/2*d*x)^9*tan(1/2*c)^6 + 600*tan(1/2*d*x)^8*tan(1/2*c)^7 + 700*tan(1/
2*d*x)^7*tan(1/2*c)^8 + 58*tan(1/2*d*x)^5*tan(1/2*c)^10 - 150*tan(1/2*d*x)^9*tan(1/2*c)^4 - 900*tan(1/2*d*x)^8
*tan(1/2*c)^5 - 2200*tan(1/2*d*x)^7*tan(1/2*c)^6 - 2400*tan(1/2*d*x)^6*tan(1/2*c)^7 - 610*tan(1/2*d*x)^5*tan(1
/2*c)^8 - 300*tan(1/2*d*x)^4*tan(1/2*c)^9 + 20*tan(1/2*d*x)^3*tan(1/2*c)^10 + 75*tan(1/2*d*x)^9*tan(1/2*c)^2 +
600*tan(1/2*d*x)^8*tan(1/2*c)^3 + 2200*tan(1/2*d*x)^7*tan(1/2*c)^4 + 4800*tan(1/2*d*x)^6*tan(1/2*c)^5 + 5380*
tan(1/2*d*x)^5*tan(1/2*c)^6 + 2000*tan(1/2*d*x)^4*tan(1/2*c)^7 + 700*tan(1/2*d*x)^3*tan(1/2*c)^8 + 15*tan(1/2*
d*x)*tan(1/2*c)^10 - 15*tan(1/2*d*x)^9 - 150*tan(1/2*d*x)^8*tan(1/2*c) - 700*tan(1/2*d*x)^7*tan(1/2*c)^2 - 240
0*tan(1/2*d*x)^6*tan(1/2*c)^3 - 5380*tan(1/2*d*x)^5*tan(1/2*c)^4 - 5960*tan(1/2*d*x)^4*tan(1/2*c)^5 - 2200*tan
(1/2*d*x)^3*tan(1/2*c)^6 - 800*tan(1/2*d*x)^2*tan(1/2*c)^7 - 75*tan(1/2*d*x)*tan(1/2*c)^8 - 30*tan(1/2*c)^9 -
20*tan(1/2*d*x)^7 + 610*tan(1/2*d*x)^5*tan(1/2*c)^2 + 2000*tan(1/2*d*x)^4*tan(1/2*c)^3 + 2200*tan(1/2*d*x)^3*t
an(1/2*c)^4 + 320*tan(1/2*d*x)^2*tan(1/2*c)^5 + 150*tan(1/2*d*x)*tan(1/2*c)^6 - 40*tan(1/2*c)^7 - 58*tan(1/2*d
*x)^5 - 300*tan(1/2*d*x)^4*tan(1/2*c) - 700*tan(1/2*d*x)^3*tan(1/2*c)^2 - 800*tan(1/2*d*x)^2*tan(1/2*c)^3 - 15
0*tan(1/2*d*x)*tan(1/2*c)^4 - 116*tan(1/2*c)^5 - 20*tan(1/2*d*x)^3 + 75*tan(1/2*d*x)*tan(1/2*c)^2 - 40*tan(1/2
*c)^3 - 15*tan(1/2*d*x) - 30*tan(1/2*c))/((a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*
c)^3 - tan(1/2*d*x)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d
*x)*tan(1/2*c) + 1)*tan(1/2*c)^10 + 5*a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3
- tan(1/2*d*x)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x)*t
an(1/2*c) + 1)*tan(1/2*c)^8 + 10*a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan
(1/2*d*x)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x)*tan(1/
2*c) + 1)*tan(1/2*c)^6 + 10*a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*
d*x)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x)*tan(1/2*c)
+ 1)*tan(1/2*c)^4 + 5*a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*d*x)^4
- 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x)*tan(1/2*c) + 1)*t
an(1/2*c)^2 + a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*d*x)^4 - 4*tan
(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x)*tan(1/2*c) + 1))*(tan(1/2
*d*x)^2 + 1)^5*d)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,{\mathrm {tan}\left (c+d\,x\right )}^4+20\,{\mathrm {tan}\left (c+d\,x\right )}^2+15\right )}{15\,d\,{\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}^{5/2}} \]

[In]

int(1/(a + a*tan(c + d*x)^2)^(5/2),x)

[Out]

(tan(c + d*x)*(20*tan(c + d*x)^2 + 8*tan(c + d*x)^4 + 15))/(15*d*(a + a*tan(c + d*x)^2)^(5/2))