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*(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)

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Rubi [A]  time = 0.131723, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, 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 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 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 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*}

Mathematica [A]  time = 0.785572, size = 131, normalized size = 0.8 \[ \frac{105 \left (8 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))+56 \left (8 a^2-b^2\right ) \tan (c+d x) \sec ^5(c+d x)+70 \left (8 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)+105 \left (8 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)+48 b \sec ^7(c+d x) (16 a+7 b \tan (c+d x))}{2688 d} \]

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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Maple [A]  time = 0.053, size = 235, normalized size = 1.4 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+{\frac{5\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{5\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{64\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{128\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,\sin \left ( dx+c \right ){b}^{2}}{128\,d}}-{\frac{5\,{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{128\,d}}+{\frac{2\,ab}{7\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{24\,d}}+{\frac{5\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

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]

1/8/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/d*sin(d*x+c)*b^2-5/128/d*b^2*ln(sec(d*x+c)+tan(d*x+c))+2/7/d*a*b/c
os(d*x+c)^7+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))

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Maxima [A]  time = 1.298, size = 297, normalized size = 1.82 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 2.04581, size = 397, normalized size = 2.44 \begin{align*} \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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \sec ^{7}{\left (c + d x \right )}\, dx \end{align*}

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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Giac [B]  time = 1.63664, size = 590, normalized size = 3.62 \begin{align*} \text{result too large to display} \end{align*}

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