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

Optimal. Leaf size=47 \[ -\frac{a^2 \log (\cos (c+d x))}{d}+\frac{2 a b \sec (c+d x)}{d}+\frac{b^2 \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0336619, antiderivative size = 47, normalized size of antiderivative = 1., 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 = {3885, 43} \[ -\frac{a^2 \log (\cos (c+d x))}{d}+\frac{2 a b \sec (c+d x)}{d}+\frac{b^2 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3885

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0568751, size = 42, normalized size = 0.89 \[ \frac{-2 a^2 \log (\cos (c+d x))+4 a b \sec (c+d x)+b^2 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 45, normalized size = 1. \begin{align*}{\frac{{b}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+2\,{\frac{ab\sec \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.974877, size = 57, normalized size = 1.21 \begin{align*} -\frac{2 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac{4 \, a b \cos \left (d x + c\right ) + b^{2}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*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*b*cos(d*x + c) + b^2)/cos(d*x + c)^2)/d

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Fricas [A]  time = 0.844613, size = 127, normalized size = 2.7 \begin{align*} -\frac{2 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 4 \, a b \cos \left (d x + c\right ) - b^{2}}{2 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*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*b*cos(d*x + c) - b^2)/(d*cos(d*x + c)^2)

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Sympy [A]  time = 0.682281, size = 60, normalized size = 1.28 \begin{align*} \begin{cases} \frac{a^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{2 a b \sec{\left (c + d x \right )}}{d} + \frac{b^{2} \sec ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \sec{\left (c \right )}\right )^{2} \tan{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.2018, size = 258, normalized size = 5.49 \begin{align*} \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{3 \, a^{2} + 8 \, a b + \frac{6 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{8 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b^{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*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)) + (3*a^2 + 8*a*b + 6*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*a*b*(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - 4*b^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