3.287 \(\int \frac{\tan ^7(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=170 \[ \frac{\left (a^2-3 b^2\right ) \sec ^3(c+d x)}{3 b^3 d}-\frac{a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 b^4 d}+\frac{\left (-3 a^2 b^2+a^4+3 b^4\right ) \sec (c+d x)}{b^5 d}-\frac{\left (a^2-b^2\right )^3 \log (a+b \sec (c+d x))}{a b^6 d}-\frac{a \sec ^4(c+d x)}{4 b^2 d}+\frac{\log (\cos (c+d x))}{a d}+\frac{\sec ^5(c+d x)}{5 b d} \]

[Out]

Log[Cos[c + d*x]]/(a*d) - ((a^2 - b^2)^3*Log[a + b*Sec[c + d*x]])/(a*b^6*d) + ((a^4 - 3*a^2*b^2 + 3*b^4)*Sec[c
+ d*x])/(b^5*d) - (a*(a^2 - 3*b^2)*Sec[c + d*x]^2)/(2*b^4*d) + ((a^2 - 3*b^2)*Sec[c + d*x]^3)/(3*b^3*d) - (a*
Sec[c + d*x]^4)/(4*b^2*d) + Sec[c + d*x]^5/(5*b*d)

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Rubi [A]  time = 0.139207, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3885, 894} \[ \frac{\left (a^2-3 b^2\right ) \sec ^3(c+d x)}{3 b^3 d}-\frac{a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 b^4 d}+\frac{\left (-3 a^2 b^2+a^4+3 b^4\right ) \sec (c+d x)}{b^5 d}-\frac{\left (a^2-b^2\right )^3 \log (a+b \sec (c+d x))}{a b^6 d}-\frac{a \sec ^4(c+d x)}{4 b^2 d}+\frac{\log (\cos (c+d x))}{a d}+\frac{\sec ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Cos[c + d*x]]/(a*d) - ((a^2 - b^2)^3*Log[a + b*Sec[c + d*x]])/(a*b^6*d) + ((a^4 - 3*a^2*b^2 + 3*b^4)*Sec[c
+ d*x])/(b^5*d) - (a*(a^2 - 3*b^2)*Sec[c + d*x]^2)/(2*b^4*d) + ((a^2 - 3*b^2)*Sec[c + d*x]^3)/(3*b^3*d) - (a*
Sec[c + d*x]^4)/(4*b^2*d) + Sec[c + d*x]^5/(5*b*d)

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\tan ^7(c+d x)}{a+b \sec (c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^3}{x (a+x)} \, dx,x,b \sec (c+d x)\right )}{b^6 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-a^4 \left (1+\frac{3 b^2 \left (-a^2+b^2\right )}{a^4}\right )+\frac{b^6}{a x}+a \left (a^2-3 b^2\right ) x-\left (a^2-3 b^2\right ) x^2+a x^3-x^4+\frac{\left (a^2-b^2\right )^3}{a (a+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{b^6 d}\\ &=\frac{\log (\cos (c+d x))}{a d}-\frac{\left (a^2-b^2\right )^3 \log (a+b \sec (c+d x))}{a b^6 d}+\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \sec (c+d x)}{b^5 d}-\frac{a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 b^4 d}+\frac{\left (a^2-3 b^2\right ) \sec ^3(c+d x)}{3 b^3 d}-\frac{a \sec ^4(c+d x)}{4 b^2 d}+\frac{\sec ^5(c+d x)}{5 b d}\\ \end{align*}

Mathematica [B]  time = 6.17435, size = 371, normalized size = 2.18 \[ \frac{\left (a^2-3 b^2\right ) \sec ^4(c+d x) (a \cos (c+d x)+b)}{3 b^3 d (a+b \sec (c+d x))}+\frac{a \left (3 b^2-a^2\right ) \sec ^3(c+d x) (a \cos (c+d x)+b)}{2 b^4 d (a+b \sec (c+d x))}+\frac{\left (-3 a^2 b^2+a^4+3 b^4\right ) \sec ^2(c+d x) (a \cos (c+d x)+b)}{b^5 d (a+b \sec (c+d x))}+\frac{\left (-3 a^3 b^2+a^5+3 a b^4\right ) \sec (c+d x) \log (\cos (c+d x)) (a \cos (c+d x)+b)}{b^6 d (a+b \sec (c+d x))}+\frac{\left (3 a^4 b^2-3 a^2 b^4-a^6+b^6\right ) \sec (c+d x) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{a b^6 d (a+b \sec (c+d x))}-\frac{a \sec ^5(c+d x) (a \cos (c+d x)+b)}{4 b^2 d (a+b \sec (c+d x))}+\frac{\sec ^6(c+d x) (a \cos (c+d x)+b)}{5 b d (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^5 - 3*a^3*b^2 + 3*a*b^4)*(b + a*Cos[c + d*x])*Log[Cos[c + d*x]]*Sec[c + d*x])/(b^6*d*(a + b*Sec[c + d*x]))
+ ((-a^6 + 3*a^4*b^2 - 3*a^2*b^4 + b^6)*(b + a*Cos[c + d*x])*Log[b + a*Cos[c + d*x]]*Sec[c + d*x])/(a*b^6*d*(
a + b*Sec[c + d*x])) + ((a^4 - 3*a^2*b^2 + 3*b^4)*(b + a*Cos[c + d*x])*Sec[c + d*x]^2)/(b^5*d*(a + b*Sec[c + d
*x])) + (a*(-a^2 + 3*b^2)*(b + a*Cos[c + d*x])*Sec[c + d*x]^3)/(2*b^4*d*(a + b*Sec[c + d*x])) + ((a^2 - 3*b^2)
*(b + a*Cos[c + d*x])*Sec[c + d*x]^4)/(3*b^3*d*(a + b*Sec[c + d*x])) - (a*(b + a*Cos[c + d*x])*Sec[c + d*x]^5)
/(4*b^2*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d*x])*Sec[c + d*x]^6)/(5*b*d*(a + b*Sec[c + d*x]))

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Maple [A]  time = 0.057, size = 292, normalized size = 1.7 \begin{align*} -{\frac{{a}^{5}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{6}}}+3\,{\frac{{a}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{4}}}-3\,{\frac{a\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{2}}}+{\frac{\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{ad}}-{\frac{a}{4\,d{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2}}{3\,d{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{db \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{4}}{d{b}^{5}\cos \left ( dx+c \right ) }}-3\,{\frac{{a}^{2}}{d{b}^{3}\cos \left ( dx+c \right ) }}+3\,{\frac{1}{db\cos \left ( dx+c \right ) }}-{\frac{{a}^{3}}{2\,d{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a}{2\,d{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{5}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d{b}^{6}}}-3\,{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d{b}^{4}}}+3\,{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d{b}^{2}}}+{\frac{1}{5\,db \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/b^6*a^5*ln(b+a*cos(d*x+c))+3/d/b^4*a^3*ln(b+a*cos(d*x+c))-3/d/b^2*a*ln(b+a*cos(d*x+c))+1/d/a*ln(b+a*cos(d
*x+c))-1/4/d/b^2*a/cos(d*x+c)^4+1/3/d/b^3/cos(d*x+c)^3*a^2-1/d/b/cos(d*x+c)^3+1/d/b^5/cos(d*x+c)*a^4-3/d/b^3/c
os(d*x+c)*a^2+3/d/b/cos(d*x+c)-1/2/d/b^4*a^3/cos(d*x+c)^2+3/2/d/b^2*a/cos(d*x+c)^2+1/d/b^6*a^5*ln(cos(d*x+c))-
3/d/b^4*a^3*ln(cos(d*x+c))+3/d/b^2*a*ln(cos(d*x+c))+1/5/d/b/cos(d*x+c)^5

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Maxima [A]  time = 0.992441, size = 247, normalized size = 1.45 \begin{align*} \frac{\frac{60 \,{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{6}} - \frac{60 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a b^{6}} - \frac{15 \, a b^{3} \cos \left (d x + c\right ) - 60 \,{\left (a^{4} - 3 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - 12 \, b^{4} + 30 \,{\left (a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 20 \,{\left (a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}}{b^{5} \cos \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(a^5 - 3*a^3*b^2 + 3*a*b^4)*log(cos(d*x + c))/b^6 - 60*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*log(a*cos(
d*x + c) + b)/(a*b^6) - (15*a*b^3*cos(d*x + c) - 60*(a^4 - 3*a^2*b^2 + 3*b^4)*cos(d*x + c)^4 - 12*b^4 + 30*(a^
3*b - 3*a*b^3)*cos(d*x + c)^3 - 20*(a^2*b^2 - 3*b^4)*cos(d*x + c)^2)/(b^5*cos(d*x + c)^5))/d

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Fricas [A]  time = 1.25118, size = 473, normalized size = 2.78 \begin{align*} -\frac{15 \, a^{2} b^{4} \cos \left (d x + c\right ) + 60 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{5} \log \left (a \cos \left (d x + c\right ) + b\right ) - 60 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 12 \, a b^{5} - 60 \,{\left (a^{5} b - 3 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} + 30 \,{\left (a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 20 \,{\left (a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}}{60 \, a b^{6} d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*a^2*b^4*cos(d*x + c) + 60*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^5*log(a*cos(d*x + c) + b)
- 60*(a^6 - 3*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^5*log(-cos(d*x + c)) - 12*a*b^5 - 60*(a^5*b - 3*a^3*b^3 + 3*a
*b^5)*cos(d*x + c)^4 + 30*(a^4*b^2 - 3*a^2*b^4)*cos(d*x + c)^3 - 20*(a^3*b^3 - 3*a*b^5)*cos(d*x + c)^2)/(a*b^6
*d*cos(d*x + c)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 7.65217, size = 1299, normalized size = 7.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*(a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*log(abs(a + b + a*(cos(d
*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^2*b^6 - a*b^7) + 60*log(abs(-(c
os(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a - 60*(a^5 - 3*a^3*b^2 + 3*a*b^4)*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - 1))/b^6 + (137*a^5 - 120*a^4*b - 411*a^3*b^2 + 320*a^2*b^3 + 411*a*b^4 - 264*b^5 + 685*a^5*(co
s(d*x + c) - 1)/(cos(d*x + c) + 1) - 480*a^4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2175*a^3*b^2*(cos(d*x +
c) - 1)/(cos(d*x + c) + 1) + 1360*a^2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2295*a*b^4*(cos(d*x + c) -
1)/(cos(d*x + c) + 1) - 1200*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1370*a^5*(cos(d*x + c) - 1)^2/(cos(d*
x + c) + 1)^2 - 720*a^4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 4470*a^3*b^2*(cos(d*x + c) - 1)^2/(cos(d
*x + c) + 1)^2 + 2000*a^2*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 5070*a*b^4*(cos(d*x + c) - 1)^2/(cos
(d*x + c) + 1)^2 - 1920*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1370*a^5*(cos(d*x + c) - 1)^3/(cos(d*x
+ c) + 1)^3 - 480*a^4*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 4470*a^3*b^2*(cos(d*x + c) - 1)^3/(cos(d*
x + c) + 1)^3 + 1200*a^2*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 5070*a*b^4*(cos(d*x + c) - 1)^3/(cos(
d*x + c) + 1)^3 - 720*b^5*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 685*a^5*(cos(d*x + c) - 1)^4/(cos(d*x +
c) + 1)^4 - 120*a^4*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 2175*a^3*b^2*(cos(d*x + c) - 1)^4/(cos(d*x +
c) + 1)^4 + 240*a^2*b^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 2295*a*b^4*(cos(d*x + c) - 1)^4/(cos(d*x
+ c) + 1)^4 - 120*b^5*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 137*a^5*(cos(d*x + c) - 1)^5/(cos(d*x + c) +
1)^5 - 411*a^3*b^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 411*a*b^4*(cos(d*x + c) - 1)^5/(cos(d*x + c) +
1)^5)/(b^6*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^5))/d