3.98 \(\int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=66 \[ -\frac {5 \tan (c+d x)}{2 a^3 d}+\frac {7 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {\tan (c+d x) (1-\sec (c+d x))}{2 a^3 d}-\frac {x}{a^3} \]

[Out]

-x/a^3+7/2*arctanh(sin(d*x+c))/a^3/d-5/2*tan(d*x+c)/a^3/d-1/2*(1-sec(d*x+c))*tan(d*x+c)/a^3/d

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Rubi [A]  time = 0.09, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3888, 3775, 3914, 3767, 8, 3770} \[ -\frac {5 \tan (c+d x)}{2 a^3 d}+\frac {7 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {\tan (c+d x) (1-\sec (c+d x))}{2 a^3 d}-\frac {x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

-(x/a^3) + (7*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (5*Tan[c + d*x])/(2*a^3*d) - ((1 - Sec[c + d*x])*Tan[c + d*x]
)/(2*a^3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3775

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 3888

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

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rubi steps

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

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Mathematica [B]  time = 0.96, size = 241, normalized size = 3.65 \[ \frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-\frac {12 \sin (d x)}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {14 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {14 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-4 x\right )}{a^3 (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]^6*Sec[c + d*x]^3*(-4*x - (14*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (14*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]])/d + 1/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - 1/(d*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^2) - (12*Sin[d*x])/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(a^3*(1 + Sec[c + d*x])^3)

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fricas [A]  time = 0.62, size = 87, normalized size = 1.32 \[ -\frac {4 \, d x \cos \left (d x + c\right )^{2} - 7 \, \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{4 \, a^{3} d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(4*d*x*cos(d*x + c)^2 - 7*cos(d*x + c)^2*log(sin(d*x + c) + 1) + 7*cos(d*x + c)^2*log(-sin(d*x + c) + 1)
+ 2*(6*cos(d*x + c) - 1)*sin(d*x + c))/(a^3*d*cos(d*x + c)^2)

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giac [A]  time = 7.54, size = 97, normalized size = 1.47 \[ -\frac {\frac {2 \, {\left (d x + c\right )}}{a^{3}} - \frac {7 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {7 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {2 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(d*x + c)/a^3 - 7*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 7*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 -
2*(7*tan(1/2*d*x + 1/2*c)^3 - 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3))/d

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maple [B]  time = 0.60, size = 144, normalized size = 2.18 \[ \frac {1}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}-\frac {1}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {7}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+a*sec(d*x+c))^3,x)

[Out]

1/2/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2+7/2/a^3/d/(tan(1/2*d*x+1/2*c)-1)-7/2/a^3/d*ln(tan(1/2*d*x+1/2*c)-1)-1/2/a^3
/d/(tan(1/2*d*x+1/2*c)+1)^2+7/2/a^3/d/(tan(1/2*d*x+1/2*c)+1)+7/2/a^3/d*ln(tan(1/2*d*x+1/2*c)+1)-2/d/a^3*arctan
(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.45, size = 171, normalized size = 2.59 \[ -\frac {\frac {2 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {4 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {7 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {7 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(2*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + 4*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/
a^3 - 7*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 7*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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mupad [B]  time = 1.31, size = 92, normalized size = 1.39 \[ \frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {x}{a^3}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + a/cos(c + d*x))^3,x)

[Out]

(7*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - x/a^3 - (5*tan(c/2 + (d*x)/2) - 7*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/
2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**6/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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