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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 170, 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} \[ \frac {\left (a^2-3 b^2\right ) \sec ^3(c+d x)}{3 b^3 d}-\frac {a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 b^4 d}+\frac {\left (-3 a^2 b^2+a^4+3 b^4\right ) \sec (c+d x)}{b^5 d}-\frac {\left (a^2-b^2\right )^3 \log (a+b \sec (c+d x))}{a b^6 d}-\frac {a \sec ^4(c+d x)}{4 b^2 d}+\frac {\log (\cos (c+d x))}{a d}+\frac {\sec ^5(c+d x)}{5 b d} \]

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 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 ^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 = 6.19, size = 371, normalized size = 2.18 \[ \frac {a \left (3 b^2-a^2\right ) \sec ^3(c+d x) (a \cos (c+d x)+b)}{2 b^4 d (a+b \sec (c+d x))}+\frac {\left (a^2-3 b^2\right ) \sec ^4(c+d x) (a \cos (c+d x)+b)}{3 b^3 d (a+b \sec (c+d x))}+\frac {\left (a^5-3 a^3 b^2+3 a b^4\right ) \sec (c+d x) \log (\cos (c+d x)) (a \cos (c+d x)+b)}{b^6 d (a+b \sec (c+d x))}+\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \sec ^2(c+d x) (a \cos (c+d x)+b)}{b^5 d (a+b \sec (c+d x))}+\frac {\left (-a^6+3 a^4 b^2-3 a^2 b^4+b^6\right ) \sec (c+d x) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{a b^6 d (a+b \sec (c+d x))}-\frac {a \sec ^5(c+d x) (a \cos (c+d x)+b)}{4 b^2 d (a+b \sec (c+d x))}+\frac {\sec ^6(c+d x) (a \cos (c+d x)+b)}{5 b d (a+b \sec (c+d x))} \]

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]))

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 205, normalized size = 1.21 \[ -\frac {15 \, a^{2} b^{4} \cos \left (d x + c\right ) + 60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{5} \log \left (a \cos \left (d x + c\right ) + b\right ) - 60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 12 \, a b^{5} - 60 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} + 30 \, {\left (a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 20 \, {\left (a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}}{60 \, a b^{6} d \cos \left (d x + c\right )^{5}} \]

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)

________________________________________________________________________________________

giac [B]  time = 11.52, size = 1052, normalized size = 6.19 \[ \text {result too large to display} \]

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

________________________________________________________________________________________

maple [A]  time = 0.48, size = 292, normalized size = 1.72 \[ -\frac {a^{5} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \,b^{6}}+\frac {3 a^{3} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \,b^{4}}-\frac {3 a \ln \left (b +a \cos \left (d x +c \right )\right )}{d \,b^{2}}+\frac {\ln \left (b +a \cos \left (d x +c \right )\right )}{d a}-\frac {a}{4 d \,b^{2} \cos \left (d x +c \right )^{4}}+\frac {a^{2}}{3 d \,b^{3} \cos \left (d x +c \right )^{3}}-\frac {1}{d b \cos \left (d x +c \right )^{3}}+\frac {a^{4}}{d \,b^{5} \cos \left (d x +c \right )}-\frac {3 a^{2}}{d \,b^{3} \cos \left (d x +c \right )}+\frac {3}{d b \cos \left (d x +c \right )}-\frac {a^{3}}{2 d \,b^{4} \cos \left (d x +c \right )^{2}}+\frac {3 a}{2 d \,b^{2} \cos \left (d x +c \right )^{2}}+\frac {a^{5} \ln \left (\cos \left (d x +c \right )\right )}{d \,b^{6}}-\frac {3 a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d \,b^{4}}+\frac {3 a \ln \left (\cos \left (d x +c \right )\right )}{d \,b^{2}}+\frac {1}{5 d b \cos \left (d x +c \right )^{5}} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 183, normalized size = 1.08 \[ \frac {\frac {60 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{6}} - \frac {60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a b^{6}} - \frac {15 \, a b^{3} \cos \left (d x + c\right ) - 60 \, {\left (a^{4} - 3 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - 12 \, b^{4} + 30 \, {\left (a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 20 \, {\left (a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}}{b^{5} \cos \left (d x + c\right )^{5}}}{60 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 2.40, size = 395, normalized size = 2.32 \[ \frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (a^4-3\,a^2\,b^2+3\,b^4\right )}{b^6\,d}-\frac {\frac {2\,\left (15\,a^4-40\,a^2\,b^2+33\,b^4\right )}{15\,b^5}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a^4+a^3\,b-2\,a^2\,b^2-2\,a\,b^3+b^4\right )}{b^5}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^4+3\,a^3\,b-10\,a^2\,b^2-8\,a\,b^3+6\,b^4\right )}{b^5}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^4+3\,a^3\,b-34\,a^2\,b^2-6\,a\,b^3+30\,b^4\right )}{3\,b^5}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (18\,a^4+9\,a^3\,b-50\,a^2\,b^2-24\,a\,b^3+48\,b^4\right )}{3\,b^5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,{\left (a^2-b^2\right )}^3}{a\,b^6\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*log(tan(c/2 + (d*x)/2)^2 - 1)*(a^4 + 3*b^4 - 3*a^2*b^2))/(b^6*d) - ((2*(15*a^4 + 33*b^4 - 40*a^2*b^2))/(15*
b^5) + (2*tan(c/2 + (d*x)/2)^8*(a^3*b - 2*a*b^3 + a^4 + b^4 - 2*a^2*b^2))/b^5 - (2*tan(c/2 + (d*x)/2)^6*(3*a^3
*b - 8*a*b^3 + 4*a^4 + 6*b^4 - 10*a^2*b^2))/b^5 - (2*tan(c/2 + (d*x)/2)^2*(3*a^3*b - 6*a*b^3 + 12*a^4 + 30*b^4
- 34*a^2*b^2))/(3*b^5) + (2*tan(c/2 + (d*x)/2)^4*(9*a^3*b - 24*a*b^3 + 18*a^4 + 48*b^4 - 50*a^2*b^2))/(3*b^5)
)/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + ta
n(c/2 + (d*x)/2)^10 - 1)) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(
c/2 + (d*x)/2)^2)*(a^2 - b^2)^3)/(a*b^6*d)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________