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

Optimal. Leaf size=57 \[ \frac {a \sec ^3(c+d x)}{3 d}+\frac {a \sec ^2(c+d x)}{2 d}-\frac {a \sec (c+d x)}{d}+\frac {a \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, 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, 75} \[ \frac {a \sec ^3(c+d x)}{3 d}+\frac {a \sec ^2(c+d x)}{2 d}-\frac {a \sec (c+d x)}{d}+\frac {a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 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)) \tan ^3(c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(a-a x) (a+a x)^2}{x^4} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a^3}{x^4}+\frac {a^3}{x^3}-\frac {a^3}{x^2}-\frac {a^3}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac {a \log (\cos (c+d x))}{d}-\frac {a \sec (c+d x)}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 55, normalized size = 0.96 \[ \frac {a \sec ^3(c+d x)}{3 d}-\frac {a \sec (c+d x)}{d}+\frac {a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 57, normalized size = 1.00 \[ \frac {6 \, a \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 2 \, a}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.96, size = 155, normalized size = 2.72 \[ -\frac {6 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 6 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {19 \, a + \frac {69 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {45 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {11 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 6*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 1)) + (19*a + 69*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 45*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1
)^2 + 11*a*(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 = 0.73, size = 104, normalized size = 1.82 \[ \frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}-\frac {a \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{3 d}-\frac {2 a \cos \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 50, normalized size = 0.88 \[ \frac {6 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {6 \, a \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, a}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.99, size = 96, normalized size = 1.68 \[ \frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {4\,a}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.92, size = 76, normalized size = 1.33 \[ \begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{3 d} + \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {2 a \sec {\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right ) \tan ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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