\(\int \frac {1}{(a+a \tan ^2(c+d x))^{7/2}} \, dx\) [286]

Optimal result

Integrand size = 16, antiderivative size = 118 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \]

[Out]

1/7*tan(d*x+c)/d/(a*sec(d*x+c)^2)^(7/2)+6/35*tan(d*x+c)/a/d/(a*sec(d*x+c)^2)^(5/2)+8/35*tan(d*x+c)/a^2/d/(a*se
c(d*x+c)^2)^(3/2)+16/35*tan(d*x+c)/a^3/d/(a*sec(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3738, 4207, 198, 197} \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}} \]

[In]

Int[(a + a*Tan[c + d*x]^2)^(-7/2),x]

[Out]

Tan[c + d*x]/(7*d*(a*Sec[c + d*x]^2)^(7/2)) + (6*Tan[c + d*x])/(35*a*d*(a*Sec[c + d*x]^2)^(5/2)) + (8*Tan[c +
d*x])/(35*a^2*d*(a*Sec[c + d*x]^2)^(3/2)) + (16*Tan[c + d*x])/(35*a^3*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 )^{7/2}} \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{9/2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{7/2}} \, dx,x,\tan (c+d x)\right )}{7 d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {24 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{35 a d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{35 a^2 d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {\left (35-35 \sin ^2(c+d x)+21 \sin ^4(c+d x)-5 \sin ^6(c+d x)\right ) \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \]

[In]

Integrate[(a + a*Tan[c + d*x]^2)^(-7/2),x]

[Out]

((35 - 35*Sin[c + d*x]^2 + 21*Sin[c + d*x]^4 - 5*Sin[c + d*x]^6)*Tan[c + d*x])/(35*a^3*d*Sqrt[a*Sec[c + d*x]^2
])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a \left (\frac {\tan \left (d x +c \right )}{7 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {7}{2}}}+\frac {\frac {6 \tan \left (d x +c \right )}{35 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}\right )}{7 a}}{a}\right )}{d}\)	\(119\)
default	\(\frac {a \left (\frac {\tan \left (d x +c \right )}{7 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {7}{2}}}+\frac {\frac {6 \tan \left (d x +c \right )}{35 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}\right )}{7 a}}{a}\right )}{d}\)	\(119\)
risch	\(-\frac {i {\mathrm e}^{8 i \left (d x +c \right )}}{896 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}-\frac {35 i {\mathrm e}^{2 i \left (d x +c \right )}}{128 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}+\frac {35 i}{128 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {7 i {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}-\frac {11 i \cos \left (6 d x +6 c \right )}{1120 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {27 \sin \left (6 d x +6 c \right )}{2240 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {7 i \cos \left (4 d x +4 c \right )}{160 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {21 \sin \left (4 d x +4 c \right )}{320 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\)	\(447\)

[In]

int(1/(a+a*tan(d*x+c)^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/d*a*(1/7/a*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(7/2)+6/7/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 = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {{\left (16 \, \tan \left (d x + c\right )^{7} + 56 \, \tan \left (d x + c\right )^{5} + 70 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{35 \, {\left (a^{4} d \tan \left (d x + c\right )^{8} + 4 \, a^{4} d \tan \left (d x + c\right )^{6} + 6 \, a^{4} d \tan \left (d x + c\right )^{4} + 4 \, a^{4} d \tan \left (d x + c\right )^{2} + a^{4} d\right )}} \]

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(7/2),x, algorithm="fricas")

[Out]

1/35*(16*tan(d*x + c)^7 + 56*tan(d*x + c)^5 + 70*tan(d*x + c)^3 + 35*tan(d*x + c))*sqrt(a*tan(d*x + c)^2 + a)/
(a^4*d*tan(d*x + c)^8 + 4*a^4*d*tan(d*x + c)^6 + 6*a^4*d*tan(d*x + c)^4 + 4*a^4*d*tan(d*x + c)^2 + a^4*d)

Sympy [F]

\[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(1/(a+a*tan(d*x+c)**2)**(7/2),x)

[Out]

Integral((a*tan(c + d*x)**2 + a)**(-7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {5 \, \sin \left (7 \, d x + 7 \, c\right ) + 49 \, \sin \left (5 \, d x + 5 \, c\right ) + 245 \, \sin \left (3 \, d x + 3 \, c\right ) + 1225 \, \sin \left (d x + c\right )}{2240 \, a^{\frac {7}{2}} d} \]

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(7/2),x, algorithm="maxima")

[Out]

1/2240*(5*sin(7*d*x + 7*c) + 49*sin(5*d*x + 5*c) + 245*sin(3*d*x + 3*c) + 1225*sin(d*x + c))/(a^(7/2)*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2158 vs. \(2 (102) = 204\).

Time = 2.85 (sec) , antiderivative size = 2158, normalized size of antiderivative = 18.29 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(7/2),x, algorithm="giac")

[Out]

-2/35*(35*tan(1/2*d*x)^13*tan(1/2*c)^14 - 245*tan(1/2*d*x)^13*tan(1/2*c)^12 - 490*tan(1/2*d*x)^12*tan(1/2*c)^1
3 + 70*tan(1/2*d*x)^11*tan(1/2*c)^14 + 735*tan(1/2*d*x)^13*tan(1/2*c)^10 + 2940*tan(1/2*d*x)^12*tan(1/2*c)^11
+ 3430*tan(1/2*d*x)^11*tan(1/2*c)^12 + 301*tan(1/2*d*x)^9*tan(1/2*c)^14 - 1225*tan(1/2*d*x)^13*tan(1/2*c)^8 -
7350*tan(1/2*d*x)^12*tan(1/2*c)^9 - 18130*tan(1/2*d*x)^11*tan(1/2*c)^10 - 19600*tan(1/2*d*x)^10*tan(1/2*c)^11
- 5243*tan(1/2*d*x)^9*tan(1/2*c)^12 - 2450*tan(1/2*d*x)^8*tan(1/2*c)^13 + 212*tan(1/2*d*x)^7*tan(1/2*c)^14 + 1
225*tan(1/2*d*x)^13*tan(1/2*c)^6 + 9800*tan(1/2*d*x)^12*tan(1/2*c)^7 + 36750*tan(1/2*d*x)^11*tan(1/2*c)^8 + 78
400*tan(1/2*d*x)^10*tan(1/2*c)^9 + 84721*tan(1/2*d*x)^9*tan(1/2*c)^10 + 34300*tan(1/2*d*x)^8*tan(1/2*c)^11 + 1
1284*tan(1/2*d*x)^7*tan(1/2*c)^12 + 301*tan(1/2*d*x)^5*tan(1/2*c)^14 - 735*tan(1/2*d*x)^13*tan(1/2*c)^4 - 7350
*tan(1/2*d*x)^12*tan(1/2*c)^5 - 36750*tan(1/2*d*x)^11*tan(1/2*c)^6 - 117600*tan(1/2*d*x)^10*tan(1/2*c)^7 - 230
055*tan(1/2*d*x)^9*tan(1/2*c)^8 - 240590*tan(1/2*d*x)^8*tan(1/2*c)^9 - 113148*tan(1/2*d*x)^7*tan(1/2*c)^10 - 3
9200*tan(1/2*d*x)^6*tan(1/2*c)^11 - 5243*tan(1/2*d*x)^5*tan(1/2*c)^12 - 1470*tan(1/2*d*x)^4*tan(1/2*c)^13 + 70
*tan(1/2*d*x)^3*tan(1/2*c)^14 + 245*tan(1/2*d*x)^13*tan(1/2*c)^2 + 2940*tan(1/2*d*x)^12*tan(1/2*c)^3 + 18130*t
an(1/2*d*x)^11*tan(1/2*c)^4 + 78400*tan(1/2*d*x)^10*tan(1/2*c)^5 + 230055*tan(1/2*d*x)^9*tan(1/2*c)^6 + 417480
*tan(1/2*d*x)^8*tan(1/2*c)^7 + 424900*tan(1/2*d*x)^7*tan(1/2*c)^8 + 219520*tan(1/2*d*x)^6*tan(1/2*c)^9 + 84721
*tan(1/2*d*x)^5*tan(1/2*c)^10 + 16660*tan(1/2*d*x)^4*tan(1/2*c)^11 + 3430*tan(1/2*d*x)^3*tan(1/2*c)^12 + 35*ta
n(1/2*d*x)*tan(1/2*c)^14 - 35*tan(1/2*d*x)^13 - 490*tan(1/2*d*x)^12*tan(1/2*c) - 3430*tan(1/2*d*x)^11*tan(1/2*
c)^2 - 19600*tan(1/2*d*x)^10*tan(1/2*c)^3 - 84721*tan(1/2*d*x)^9*tan(1/2*c)^4 - 240590*tan(1/2*d*x)^8*tan(1/2*
c)^5 - 424900*tan(1/2*d*x)^7*tan(1/2*c)^6 - 432320*tan(1/2*d*x)^6*tan(1/2*c)^7 - 230055*tan(1/2*d*x)^5*tan(1/2
*c)^8 - 91042*tan(1/2*d*x)^4*tan(1/2*c)^9 - 18130*tan(1/2*d*x)^3*tan(1/2*c)^10 - 3920*tan(1/2*d*x)^2*tan(1/2*c
)^11 - 245*tan(1/2*d*x)*tan(1/2*c)^12 - 70*tan(1/2*c)^13 - 70*tan(1/2*d*x)^11 + 5243*tan(1/2*d*x)^9*tan(1/2*c)
^2 + 34300*tan(1/2*d*x)^8*tan(1/2*c)^3 + 113148*tan(1/2*d*x)^7*tan(1/2*c)^4 + 219520*tan(1/2*d*x)^6*tan(1/2*c)
^5 + 230055*tan(1/2*d*x)^5*tan(1/2*c)^6 + 108696*tan(1/2*d*x)^4*tan(1/2*c)^7 + 36750*tan(1/2*d*x)^3*tan(1/2*c)
^8 + 3136*tan(1/2*d*x)^2*tan(1/2*c)^9 + 735*tan(1/2*d*x)*tan(1/2*c)^10 - 140*tan(1/2*c)^11 - 301*tan(1/2*d*x)^
9 - 2450*tan(1/2*d*x)^8*tan(1/2*c) - 11284*tan(1/2*d*x)^7*tan(1/2*c)^2 - 39200*tan(1/2*d*x)^6*tan(1/2*c)^3 - 8
4721*tan(1/2*d*x)^5*tan(1/2*c)^4 - 91042*tan(1/2*d*x)^4*tan(1/2*c)^5 - 36750*tan(1/2*d*x)^3*tan(1/2*c)^6 - 127
68*tan(1/2*d*x)^2*tan(1/2*c)^7 - 1225*tan(1/2*d*x)*tan(1/2*c)^8 - 602*tan(1/2*c)^9 - 212*tan(1/2*d*x)^7 + 5243
*tan(1/2*d*x)^5*tan(1/2*c)^2 + 16660*tan(1/2*d*x)^4*tan(1/2*c)^3 + 18130*tan(1/2*d*x)^3*tan(1/2*c)^4 + 3136*ta
n(1/2*d*x)^2*tan(1/2*c)^5 + 1225*tan(1/2*d*x)*tan(1/2*c)^6 - 424*tan(1/2*c)^7 - 301*tan(1/2*d*x)^5 - 1470*tan(
1/2*d*x)^4*tan(1/2*c) - 3430*tan(1/2*d*x)^3*tan(1/2*c)^2 - 3920*tan(1/2*d*x)^2*tan(1/2*c)^3 - 735*tan(1/2*d*x)
*tan(1/2*c)^4 - 602*tan(1/2*c)^5 - 70*tan(1/2*d*x)^3 + 245*tan(1/2*d*x)*tan(1/2*c)^2 - 140*tan(1/2*c)^3 - 35*t
an(1/2*d*x) - 70*tan(1/2*c))/((a^(7/2)*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)^14 + 7*a^(7/2)*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)^12 + 21*a^(7/2)*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)^10 + 35*a^(7/2)*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)^8 + 35*a^(7/2)*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 + 21*a^(7/2)*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 +
7*a^(7/2)*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)^2 + a^(7/2
)*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)^7*d)

Mupad [B] (verification not implemented)

Time = 11.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {16\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}+\frac {8\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^2}+\frac {6\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{7\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^4} \]

[In]

int(1/(a + a*tan(c + d*x)^2)^(7/2),x)

[Out]

(16*tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(35*a^4*d*(tan(c + d*x)^2 + 1)) + (8*tan(c + d*x)*(a + a*tan(c
+ d*x)^2)^(1/2))/(35*a^4*d*(tan(c + d*x)^2 + 1)^2) + (6*tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(35*a^4*d*(
tan(c + d*x)^2 + 1)^3) + (tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(7*a^4*d*(tan(c + d*x)^2 + 1)^4)