3.23 \(\int (a+a \sec (c+d x))^2 \tan (c+d x) \, dx\)

Optimal. Leaf size=48 \[ \frac {a^2 \sec ^2(c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, 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} \[ \frac {a^2 \sec ^2(c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Tan[c + d*x],x]

[Out]

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

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Mathematica [A]  time = 0.15, size = 51, normalized size = 1.06 \[ -\frac {a^2 \sec ^2(c+d x) (-4 \cos (c+d x)+\cos (2 (c+d x)) \log (\cos (c+d x))+\log (\cos (c+d x))-1)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Tan[c + d*x],x]

[Out]

-1/2*(a^2*(-1 - 4*Cos[c + d*x] + Log[Cos[c + d*x]] + Cos[2*(c + d*x)]*Log[Cos[c + d*x]])*Sec[c + d*x]^2)/d

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fricas [A]  time = 0.62, size = 52, normalized size = 1.08 \[ -\frac {2 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 4 \, a^{2} \cos \left (d x + c\right ) - a^{2}}{2 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.39, size = 142, normalized size = 2.96 \[ \frac {2 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {11 \, a^{2} + \frac {10 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c),x, algorithm="giac")

[Out]

1/2*(2*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 2*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (11*a^2 + 10*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^2*(cos(d*x + c) - 1)^2/(cos(d*x
+ c) + 1)^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d

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maple [A]  time = 0.30, size = 46, normalized size = 0.96 \[ \frac {a^{2} \left (\sec ^{2}\left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} \sec \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*tan(d*x+c),x)

[Out]

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

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maxima [A]  time = 0.32, size = 43, normalized size = 0.90 \[ -\frac {2 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {4 \, a^{2} \cos \left (d x + c\right ) + a^{2}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.21, size = 76, normalized size = 1.58 \[ \frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,a^2}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)*(a + a/cos(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 0.48, size = 60, normalized size = 1.25 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \sec ^{2}{\left (c + d x \right )}}{2 d} + \frac {2 a^{2} \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right )^{2} \tan {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*tan(d*x+c),x)

[Out]

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

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