3.41 \(\int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx\)

Optimal. Leaf size=66 \[ \frac {a^3 \sec ^3(c+d x)}{3 d}+\frac {3 a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}-\frac {a^3 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3879, 43} \[ \frac {a^3 \sec ^3(c+d x)}{3 d}+\frac {3 a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}-\frac {a^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.16, size = 64, normalized size = 0.97 \[ -\frac {a^3 \sec ^3(c+d x) (-18 \cos (2 (c+d x))+9 \cos (c+d x) (\log (\cos (c+d x))-2)+3 \cos (3 (c+d x)) \log (\cos (c+d x))-22)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(a^3*(-22 - 18*Cos[2*(c + d*x)] + 9*Cos[c + d*x]*(-2 + Log[Cos[c + d*x]]) + 3*Cos[3*(c + d*x)]*Log[Cos[c
+ d*x]])*Sec[c + d*x]^3)/d

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fricas [A]  time = 0.66, size = 65, normalized size = 0.98 \[ -\frac {6 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 18 \, a^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a^3*cos(d*x + c)^3*log(-cos(d*x + c)) - 18*a^3*cos(d*x + c)^2 - 9*a^3*cos(d*x + c) - 2*a^3)/(d*cos(d*x
+ c)^3)

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giac [B]  time = 0.51, size = 167, normalized size = 2.53 \[ \frac {6 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {51 \, a^{3} + \frac {69 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {45 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {11 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 6*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (51*a^3 + 69*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 45*a^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)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3
)/d

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maple [A]  time = 0.30, size = 62, normalized size = 0.94 \[ \frac {a^{3} \left (\sec ^{3}\left (d x +c \right )\right )}{3 d}+\frac {3 a^{3} \left (\sec ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \sec \left (d x +c \right )}{d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^3*sec(d*x+c)^3/d+3/2*a^3*sec(d*x+c)^2/d+3*a^3*sec(d*x+c)/d+a^3/d*ln(sec(d*x+c))

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maxima [A]  time = 0.68, size = 58, normalized size = 0.88 \[ -\frac {6 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac {18 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a^3*log(cos(d*x + c)) - (18*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 2*a^3)/cos(d*x + c)^3)/d

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mupad [B]  time = 1.94, size = 105, normalized size = 1.59 \[ \frac {2\,a^3\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {20\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^3*atanh(tan(c/2 + (d*x)/2)^2))/d - (2*a^3*tan(c/2 + (d*x)/2)^4 - 6*a^3*tan(c/2 + (d*x)/2)^2 + (20*a^3)/3)
/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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sympy [A]  time = 0.92, size = 76, normalized size = 1.15 \[ \begin {cases} \frac {a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{3} \sec ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 a^{3} \sec ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 a^{3} \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right )^{3} \tan {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*log(tan(c + d*x)**2 + 1)/(2*d) + a**3*sec(c + d*x)**3/(3*d) + 3*a**3*sec(c + d*x)**2/(2*d) + 3
*a**3*sec(c + d*x)/d, Ne(d, 0)), (x*(a*sec(c) + a)**3*tan(c), True))

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