3.469 \(\int (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=73 \[ \frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 x+\frac{5 a b^2 \tan (c+d x)}{2 d}+\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

[Out]

a^3*x + (b*(6*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a*b^2*Tan[c + d*x])/(2*d) + (b^2*(a + b*Sec[c + d*x
])*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.0485526, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3782, 3770, 3767, 8} \[ \frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 x+\frac{5 a b^2 \tan (c+d x)}{2 d}+\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^3,x]

[Out]

a^3*x + (b*(6*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a*b^2*Tan[c + d*x])/(2*d) + (b^2*(a + b*Sec[c + d*x
])*Tan[c + d*x])/(2*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \sec (c+d x))^3 \, dx &=\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (2 a^3+b \left (6 a^2+b^2\right ) \sec (c+d x)+5 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=a^3 x+\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac{1}{2} \left (5 a b^2\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (b \left (6 a^2+b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=a^3 x+\frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{\left (5 a b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 x+\frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a b^2 \tan (c+d x)}{2 d}+\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.154597, size = 55, normalized size = 0.75 \[ \frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))+2 a^3 d x+b^2 \tan (c+d x) (6 a+b \sec (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^3,x]

[Out]

(2*a^3*d*x + b*(6*a^2 + b^2)*ArcTanh[Sin[c + d*x]] + b^2*(6*a + b*Sec[c + d*x])*Tan[c + d*x])/(2*d)

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Maple [A]  time = 0.027, size = 95, normalized size = 1.3 \begin{align*}{a}^{3}x+{\frac{{a}^{3}c}{d}}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20637, size = 126, normalized size = 1.73 \begin{align*} a^{3} x - \frac{b^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{4 \, d} + \frac{3 \, a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac{3 \, a b^{2} \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 1/4*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1))/d + 3*a^
2*b*log(sec(d*x + c) + tan(d*x + c))/d + 3*a*b^2*tan(d*x + c)/d

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Fricas [A]  time = 1.71792, size = 281, normalized size = 3.85 \begin{align*} \frac{4 \, a^{3} d x \cos \left (d x + c\right )^{2} +{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, a b^{2} \cos \left (d x + c\right ) + b^{3}\right )} \sin \left (d x + c\right )}{4 \, 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))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x))**3, x)

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Giac [B]  time = 1.26545, size = 196, normalized size = 2.68 \begin{align*} \frac{2 \,{\left (d x + c\right )} a^{3} +{\left (6 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (6 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 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))^3,x, algorithm="giac")

[Out]

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