3.89 \(\int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=46 \[ \frac {\sec (c+d x)}{a^3 d}+\frac {3 \log (\cos (c+d x))}{a^3 d}-\frac {4 \log (\cos (c+d x)+1)}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 46, 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 = {3879, 88} \[ \frac {\sec (c+d x)}{a^3 d}+\frac {3 \log (\cos (c+d x))}{a^3 d}-\frac {4 \log (\cos (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(3*Log[Cos[c + d*x]])/(a^3*d) - (4*Log[1 + Cos[c + d*x]])/(a^3*d) + Sec[c + d*x]/(a^3*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3879

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 36, normalized size = 0.78 \[ \frac {\sec (c+d x)-8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 \log (\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(-8*Log[Cos[(c + d*x)/2]] + 3*Log[Cos[c + d*x]] + Sec[c + d*x])/(a^3*d)

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fricas [A]  time = 0.66, size = 53, normalized size = 1.15 \[ \frac {3 \, \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 4 \, \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 1}{a^{3} d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 5.49, size = 112, normalized size = 2.43 \[ \frac {\frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} + \frac {3 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{3}} - \frac {\frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 1}{a^{3} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

(log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 + 3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) -
1))/a^3 - (3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)/(a^3*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)))/d

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maple [A]  time = 0.52, size = 46, normalized size = 1.00 \[ \frac {\sec \left (d x +c \right )}{a^{3} d}+\frac {\ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}}-\frac {4 \ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+a*sec(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.33, size = 45, normalized size = 0.98 \[ -\frac {\frac {4 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {3 \, \log \left (\cos \left (d x + c\right )\right )}{a^{3}} - \frac {1}{a^{3} \cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.27, size = 72, normalized size = 1.57 \[ \frac {3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{a^3\,d}-\frac {2}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + a/cos(c + d*x))^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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