3.288 \(\int \frac{\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=108 \[ \frac{\left (a^2-2 b^2\right ) \sec (c+d x)}{b^3 d}-\frac{\left (a^2-b^2\right )^2 \log (a+b \sec (c+d x))}{a b^4 d}-\frac{a \sec ^2(c+d x)}{2 b^2 d}-\frac{\log (\cos (c+d x))}{a d}+\frac{\sec ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.0956663, antiderivative size = 108, 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} \[ \frac{\left (a^2-2 b^2\right ) \sec (c+d x)}{b^3 d}-\frac{\left (a^2-b^2\right )^2 \log (a+b \sec (c+d x))}{a b^4 d}-\frac{a \sec ^2(c+d x)}{2 b^2 d}-\frac{\log (\cos (c+d x))}{a d}+\frac{\sec ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

-(Log[Cos[c + d*x]]/(a*d)) - ((a^2 - b^2)^2*Log[a + b*Sec[c + d*x]])/(a*b^4*d) + ((a^2 - 2*b^2)*Sec[c + d*x])/
(b^3*d) - (a*Sec[c + d*x]^2)/(2*b^2*d) + Sec[c + d*x]^3/(3*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 ^5(c+d x)}{a+b \sec (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{x (a+x)} \, dx,x,b \sec (c+d x)\right )}{b^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2 \left (1-\frac{2 b^2}{a^2}\right )+\frac{b^4}{a x}-a x+x^2-\frac{\left (a^2-b^2\right )^2}{a (a+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{b^4 d}\\ &=-\frac{\log (\cos (c+d x))}{a d}-\frac{\left (a^2-b^2\right )^2 \log (a+b \sec (c+d x))}{a b^4 d}+\frac{\left (a^2-2 b^2\right ) \sec (c+d x)}{b^3 d}-\frac{a \sec ^2(c+d x)}{2 b^2 d}+\frac{\sec ^3(c+d x)}{3 b d}\\ \end{align*}

Mathematica [A]  time = 0.368648, size = 108, normalized size = 1. \[ \frac{-3 a^2 b^2 \sec ^2(c+d x)+6 a b \left (a^2-2 b^2\right ) \sec (c+d x)+6 a^2 \left (a^2-2 b^2\right ) \log (\cos (c+d x))-6 \left (a^2-b^2\right )^2 \log (a \cos (c+d x)+b)+2 a b^3 \sec ^3(c+d x)}{6 a b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.052, size = 163, normalized size = 1.5 \begin{align*} -{\frac{{a}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{4}}}+2\,{\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}{2\,d{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}}{d{b}^{3}\cos \left ( dx+c \right ) }}-2\,{\frac{1}{db\cos \left ( dx+c \right ) }}+{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d{b}^{4}}}-2\,{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d{b}^{2}}}+{\frac{1}{3\,db \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+b*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.945829, size = 149, normalized size = 1.38 \begin{align*} \frac{\frac{6 \,{\left (a^{3} - 2 \, a b^{2}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{4}} - \frac{6 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a b^{4}} - \frac{3 \, a b \cos \left (d x + c\right ) - 6 \,{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}}{b^{3} \cos \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.12712, size = 305, normalized size = 2.82 \begin{align*} -\frac{3 \, a^{2} b^{2} \cos \left (d x + c\right ) + 6 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (a \cos \left (d x + c\right ) + b\right ) - 6 \,{\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 2 \, a b^{3} - 6 \,{\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}}{6 \, a b^{4} d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\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)**5/(a+b*sec(d*x+c)),x)

[Out]

Integral(tan(c + d*x)**5/(a + b*sec(c + d*x)), x)

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Giac [B]  time = 3.62829, size = 636, normalized size = 5.89 \begin{align*} -\frac{\frac{6 \,{\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} \log \left ({\left | a + b + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} b^{4} - a b^{5}} - \frac{6 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a} - \frac{6 \,{\left (a^{3} - 2 \, a b^{2}\right )} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b^{4}} + \frac{11 \, a^{3} - 12 \, a^{2} b - 22 \, a b^{2} + 20 \, b^{3} + \frac{33 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{24 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{78 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{48 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{33 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{12 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{78 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{12 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{11 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{22 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{b^{4}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*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^4 - a*b^5) - 6*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 1))/a - 6*(a^3 - 2*a*b^2)*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/b^4 + (11*a^3 - 12*a
^2*b - 22*a*b^2 + 20*b^3 + 33*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 24*a^2*b*(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - 78*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 48*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3
3*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 12*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 78*a*b^
2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 12*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 11*a^3*(cos(d
*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 22*a*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/(b^4*((cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 1)^3))/d