3.523 \(\int \sec ^7(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=163 \[ \frac {5 \left (8 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac {\left (8 a^2-b^2\right ) \tan (c+d x) \sec ^5(c+d x)}{48 d}+\frac {5 \left (8 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{192 d}+\frac {5 \left (8 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{128 d}+\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d} \]

[Out]

5/128*(8*a^2-b^2)*arctanh(sin(d*x+c))/d+9/56*a*b*sec(d*x+c)^7/d+5/128*(8*a^2-b^2)*sec(d*x+c)*tan(d*x+c)/d+5/19
2*(8*a^2-b^2)*sec(d*x+c)^3*tan(d*x+c)/d+1/48*(8*a^2-b^2)*sec(d*x+c)^5*tan(d*x+c)/d+1/8*b*sec(d*x+c)^7*(a+b*tan
(d*x+c))/d

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Rubi [A]  time = 0.13, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3508, 3486, 3768, 3770} \[ \frac {5 \left (8 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac {\left (8 a^2-b^2\right ) \tan (c+d x) \sec ^5(c+d x)}{48 d}+\frac {5 \left (8 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{192 d}+\frac {5 \left (8 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{128 d}+\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + b*Tan[c + d*x])^2,x]

[Out]

(5*(8*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(128*d) + (9*a*b*Sec[c + d*x]^7)/(56*d) + (5*(8*a^2 - b^2)*Sec[c + d*x
]*Tan[c + d*x])/(128*d) + (5*(8*a^2 - b^2)*Sec[c + d*x]^3*Tan[c + d*x])/(192*d) + ((8*a^2 - b^2)*Sec[c + d*x]^
5*Tan[c + d*x])/(48*d) + (b*Sec[c + d*x]^7*(a + b*Tan[c + d*x]))/(8*d)

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \sec ^7(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d}+\frac {1}{8} \int \sec ^7(c+d x) \left (8 a^2-b^2+9 a b \tan (c+d x)\right ) \, dx\\ &=\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d}+\frac {1}{8} \left (8 a^2-b^2\right ) \int \sec ^7(c+d x) \, dx\\ &=\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {\left (8 a^2-b^2\right ) \sec ^5(c+d x) \tan (c+d x)}{48 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d}+\frac {1}{48} \left (5 \left (8 a^2-b^2\right )\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {5 \left (8 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{192 d}+\frac {\left (8 a^2-b^2\right ) \sec ^5(c+d x) \tan (c+d x)}{48 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d}+\frac {1}{64} \left (5 \left (8 a^2-b^2\right )\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {5 \left (8 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{128 d}+\frac {5 \left (8 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{192 d}+\frac {\left (8 a^2-b^2\right ) \sec ^5(c+d x) \tan (c+d x)}{48 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d}+\frac {1}{128} \left (5 \left (8 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {5 \left (8 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac {9 a b \sec ^7(c+d x)}{56 d}+\frac {5 \left (8 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{128 d}+\frac {5 \left (8 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{192 d}+\frac {\left (8 a^2-b^2\right ) \sec ^5(c+d x) \tan (c+d x)}{48 d}+\frac {b \sec ^7(c+d x) (a+b \tan (c+d x))}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 131, normalized size = 0.80 \[ \frac {105 \left (8 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))+56 \left (8 a^2-b^2\right ) \tan (c+d x) \sec ^5(c+d x)+70 \left (8 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)+105 \left (8 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)+48 b \sec ^7(c+d x) (16 a+7 b \tan (c+d x))}{2688 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(105*(8*a^2 - b^2)*ArcTanh[Sin[c + d*x]] + 105*(8*a^2 - b^2)*Sec[c + d*x]*Tan[c + d*x] + 70*(8*a^2 - b^2)*Sec[
c + d*x]^3*Tan[c + d*x] + 56*(8*a^2 - b^2)*Sec[c + d*x]^5*Tan[c + d*x] + 48*b*Sec[c + d*x]^7*(16*a + 7*b*Tan[c
+ d*x]))/(2688*d)

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fricas [A]  time = 0.75, size = 163, normalized size = 1.00 \[ \frac {105 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{8} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{8} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 1536 \, a b \cos \left (d x + c\right ) + 14 \, {\left (15 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 10 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, b^{2}\right )} \sin \left (d x + c\right )}{5376 \, d \cos \left (d x + c\right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5376*(105*(8*a^2 - b^2)*cos(d*x + c)^8*log(sin(d*x + c) + 1) - 105*(8*a^2 - b^2)*cos(d*x + c)^8*log(-sin(d*x
+ c) + 1) + 1536*a*b*cos(d*x + c) + 14*(15*(8*a^2 - b^2)*cos(d*x + c)^6 + 10*(8*a^2 - b^2)*cos(d*x + c)^4 + 8
*(8*a^2 - b^2)*cos(d*x + c)^2 + 48*b^2)*sin(d*x + c))/(d*cos(d*x + c)^8)

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giac [B]  time = 0.60, size = 437, normalized size = 2.68 \[ \frac {105 \, {\left (8 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (8 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (1848 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 105 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 5376 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} - 3416 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 2779 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 5376 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 6328 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 6265 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 26880 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 4760 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 12355 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 26880 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 4760 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12355 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 16128 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6328 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6265 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16128 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3416 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2779 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 768 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1848 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 768 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{8}}}{2688 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2688*(105*(8*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(8*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(1848*a^2*tan(1/2*d*x + 1/2*c)^15 + 105*b^2*tan(1/2*d*x + 1/2*c)^15 - 5376*a*b*tan(1/2*d*x + 1/2*c)^
14 - 3416*a^2*tan(1/2*d*x + 1/2*c)^13 + 2779*b^2*tan(1/2*d*x + 1/2*c)^13 + 5376*a*b*tan(1/2*d*x + 1/2*c)^12 +
6328*a^2*tan(1/2*d*x + 1/2*c)^11 + 6265*b^2*tan(1/2*d*x + 1/2*c)^11 - 26880*a*b*tan(1/2*d*x + 1/2*c)^10 - 4760
*a^2*tan(1/2*d*x + 1/2*c)^9 + 12355*b^2*tan(1/2*d*x + 1/2*c)^9 + 26880*a*b*tan(1/2*d*x + 1/2*c)^8 - 4760*a^2*t
an(1/2*d*x + 1/2*c)^7 + 12355*b^2*tan(1/2*d*x + 1/2*c)^7 - 16128*a*b*tan(1/2*d*x + 1/2*c)^6 + 6328*a^2*tan(1/2
*d*x + 1/2*c)^5 + 6265*b^2*tan(1/2*d*x + 1/2*c)^5 + 16128*a*b*tan(1/2*d*x + 1/2*c)^4 - 3416*a^2*tan(1/2*d*x +
1/2*c)^3 + 2779*b^2*tan(1/2*d*x + 1/2*c)^3 - 768*a*b*tan(1/2*d*x + 1/2*c)^2 + 1848*a^2*tan(1/2*d*x + 1/2*c) +
105*b^2*tan(1/2*d*x + 1/2*c) + 768*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^8)/d

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maple [A]  time = 0.45, size = 235, normalized size = 1.44 \[ \frac {a^{2} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 a^{2} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{24 d}+\frac {5 a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {2 a b}{7 d \cos \left (d x +c \right )^{7}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {5 b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{6}}+\frac {5 b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}+\frac {5 b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}+\frac {5 b^{2} \sin \left (d x +c \right )}{128 d}-\frac {5 b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+b*tan(d*x+c))^2,x)

[Out]

1/6/d*a^2*tan(d*x+c)*sec(d*x+c)^5+5/24*a^2*sec(d*x+c)^3*tan(d*x+c)/d+5/16*a^2*sec(d*x+c)*tan(d*x+c)/d+5/16/d*a
^2*ln(sec(d*x+c)+tan(d*x+c))+2/7/d*a*b/cos(d*x+c)^7+1/8/d*b^2*sin(d*x+c)^3/cos(d*x+c)^8+5/48/d*b^2*sin(d*x+c)^
3/cos(d*x+c)^6+5/64/d*b^2*sin(d*x+c)^3/cos(d*x+c)^4+5/128/d*b^2*sin(d*x+c)^3/cos(d*x+c)^2+5/128*b^2*sin(d*x+c)
/d-5/128/d*b^2*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.35, size = 220, normalized size = 1.35 \[ \frac {7 \, b^{2} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 56 \, a^{2} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {1536 \, a b}{\cos \left (d x + c\right )^{7}}}{5376 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5376*(7*b^2*(2*(15*sin(d*x + c)^7 - 55*sin(d*x + c)^5 + 73*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x + c)^8
- 4*sin(d*x + c)^6 + 6*sin(d*x + c)^4 - 4*sin(d*x + c)^2 + 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c
) - 1)) - 56*a^2*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)
^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 1536*a*b/cos(d*x + c)^7)/d

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mupad [B]  time = 7.10, size = 432, normalized size = 2.65 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {5\,a^2}{8}-\frac {5\,b^2}{64}\right )}{d}+\frac {\left (\frac {11\,a^2}{8}+\frac {5\,b^2}{64}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+\left (\frac {397\,b^2}{192}-\frac {61\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (\frac {113\,a^2}{24}+\frac {895\,b^2}{192}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-20\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {1765\,b^2}{192}-\frac {85\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+20\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {1765\,b^2}{192}-\frac {85\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {113\,a^2}{24}+\frac {895\,b^2}{192}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {397\,b^2}{192}-\frac {61\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{7}+\left (\frac {11\,a^2}{8}+\frac {5\,b^2}{64}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4\,a\,b}{7}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^2/cos(c + d*x)^7,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((5*a^2)/8 - (5*b^2)/64))/d + ((4*a*b)/7 + tan(c/2 + (d*x)/2)^15*((11*a^2)/8 + (5*b
^2)/64) - tan(c/2 + (d*x)/2)^3*((61*a^2)/24 - (397*b^2)/192) - tan(c/2 + (d*x)/2)^13*((61*a^2)/24 - (397*b^2)/
192) + tan(c/2 + (d*x)/2)^5*((113*a^2)/24 + (895*b^2)/192) + tan(c/2 + (d*x)/2)^11*((113*a^2)/24 + (895*b^2)/1
92) - tan(c/2 + (d*x)/2)^7*((85*a^2)/24 - (1765*b^2)/192) - tan(c/2 + (d*x)/2)^9*((85*a^2)/24 - (1765*b^2)/192
) + tan(c/2 + (d*x)/2)*((11*a^2)/8 + (5*b^2)/64) - (4*a*b*tan(c/2 + (d*x)/2)^2)/7 + 12*a*b*tan(c/2 + (d*x)/2)^
4 - 12*a*b*tan(c/2 + (d*x)/2)^6 + 20*a*b*tan(c/2 + (d*x)/2)^8 - 20*a*b*tan(c/2 + (d*x)/2)^10 + 4*a*b*tan(c/2 +
(d*x)/2)^12 - 4*a*b*tan(c/2 + (d*x)/2)^14)/(d*(28*tan(c/2 + (d*x)/2)^4 - 8*tan(c/2 + (d*x)/2)^2 - 56*tan(c/2
+ (d*x)/2)^6 + 70*tan(c/2 + (d*x)/2)^8 - 56*tan(c/2 + (d*x)/2)^10 + 28*tan(c/2 + (d*x)/2)^12 - 8*tan(c/2 + (d*
x)/2)^14 + tan(c/2 + (d*x)/2)^16 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{7}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**7, x)

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