3.301 \(\int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=121 \[ \frac {\left (a^2-b^2\right )^2}{a b^4 d (a+b \sec (c+d x))}+\frac {\left (3 a^2+b^2\right ) \left (a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^4 d}-\frac {\log (\cos (c+d x))}{a^2 d}-\frac {2 a \sec (c+d x)}{b^3 d}+\frac {\sec ^2(c+d x)}{2 b^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 121, normalized size of antiderivative = 1.00, 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} \[ \frac {\left (a^2-b^2\right )^2}{a b^4 d (a+b \sec (c+d x))}+\frac {\left (3 a^2+b^2\right ) \left (a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^4 d}-\frac {\log (\cos (c+d x))}{a^2 d}-\frac {2 a \sec (c+d x)}{b^3 d}+\frac {\sec ^2(c+d x)}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]

[Out]

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

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]))

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.60, size = 187, normalized size = 1.55 \[ \frac {b \left (-3 a^3 b \sec (c+d x)+a^2 b^2 \sec ^2(c+d x)-2 \left (3 a^4+a^2 \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))-2 a^2 b^2+\left (-3 a^4+2 a^2 b^2+b^4\right ) \log (a \cos (c+d x)+b)+b^4\right )\right )-2 a \cos (c+d x) \left (a^2 \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))+\left (-3 a^4+2 a^2 b^2+b^4\right ) \log (a \cos (c+d x)+b)\right )}{2 a^2 b^4 d (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 219, normalized size = 1.81 \[ -\frac {3 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} + 2 \, {\left (3 \, a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left ({\left (3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{4} b - 2 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + 2 \, {\left ({\left (3 \, a^{5} - 2 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right )}{2 \, {\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 6.54, size = 568, normalized size = 4.69 \[ \frac {\frac {2 \, {\left (3 \, a^{5} - 3 \, 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^{3} b^{4} - a^{2} b^{5}} + \frac {2 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (3 \, a^{2} - 2 \, 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 {9 \, a^{2} - 8 \, a b - 6 \, b^{2} + \frac {18 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {16 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{b^{4} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}} - \frac {2 \, {\left (3 \, a^{5} + 5 \, a^{4} b - 4 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5} + \frac {3 \, a^{5} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, a^{4} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a^{3} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{2} b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b^{5} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\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 )} a^{2} b^{4}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(2*(3*a^5 - 3*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^3*b^4 - a^2*b^5) + 2*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) + 1))/a^2 - 2*(3*a^2 - 2*b^2)*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/b^4 + (9*a^2
- 8*a*b - 6*b^2 + 18*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) -
16*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6*b^2*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(b^4*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2) - 2*(3*a^5 + 5*a^4*b
- 4*a^2*b^3 - 3*a*b^4 - b^5 + 3*a^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*a^4*b*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 2*a^3*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*a^2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) - a*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((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))/d

________________________________________________________________________________________

maple [A]  time = 0.48, size = 192, normalized size = 1.59 \[ -\frac {a^{2}}{d \,b^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2}{d b \left (b +a \cos \left (d x +c \right )\right )}-\frac {b}{d \,a^{2} \left (b +a \cos \left (d x +c \right )\right )}+\frac {3 a^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \,b^{4}}-\frac {2 \ln \left (b +a \cos \left (d x +c \right )\right )}{d \,b^{2}}-\frac {\ln \left (b +a \cos \left (d x +c \right )\right )}{d \,a^{2}}-\frac {3 \ln \left (\cos \left (d x +c \right )\right ) a^{2}}{d \,b^{4}}+\frac {2 \ln \left (\cos \left (d x +c \right )\right )}{d \,b^{2}}+\frac {1}{2 d \,b^{2} \cos \left (d x +c \right )^{2}}-\frac {2 a}{d \,b^{3} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+b*sec(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 149, normalized size = 1.23 \[ -\frac {\frac {3 \, a^{3} b \cos \left (d x + c\right ) - a^{2} b^{2} + 2 \, {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}}{a^{3} b^{3} \cos \left (d x + c\right )^{3} + a^{2} b^{4} \cos \left (d x + c\right )^{2}} + \frac {2 \, {\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{4}} - \frac {2 \, {\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{4}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.13, size = 286, normalized size = 2.36 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-6\,a^3-3\,a^2\,b+a\,b^2+2\,b^3\right )}{a\,b^3}-\frac {2\,\left (-3\,a^3-3\,a^2\,b+a\,b^2+b^3\right )}{a\,b^3}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a-b\right )\,\left (3\,a^2+3\,a\,b+b^2\right )}{a\,b^3}}{d\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (3\,a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-3\,a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (3\,a^2-2\,b^2\right )}{b^4\,d}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (-3\,a^4+2\,a^2\,b^2+b^4\right )}{a^2\,b^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + b/cos(c + d*x))^2,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**5/(a + b*sec(c + d*x))**2, x)

________________________________________________________________________________________