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

Optimal. Leaf size=66 \[ \frac{5 a^3 \tan (c+d x)}{2 d}+\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+a^3 x \]

[Out]

a^3*x + (7*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*Tan[c + d*x])/(2*d) + ((a^3 + a^3*Sec[c + d*x])*Tan[c + d
*x])/(2*d)

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Rubi [A]  time = 0.0473646, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3775, 3914, 3767, 8, 3770} \[ \frac{5 a^3 \tan (c+d x)}{2 d}+\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+a^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3,x]

[Out]

a^3*x + (7*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*Tan[c + d*x])/(2*d) + ((a^3 + a^3*Sec[c + d*x])*Tan[c + d
*x])/(2*d)

Rule 3775

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[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

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]

Rule 3770

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^3 \, dx &=\frac{\left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac{1}{2} a \int (a+a \sec (c+d x)) (2 a+5 a \sec (c+d x)) \, dx\\ &=a^3 x+\frac{\left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac{1}{2} \left (5 a^3\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (7 a^3\right ) \int \sec (c+d x) \, dx\\ &=a^3 x+\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{2 d}-\frac{\left (5 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 x+\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a^3 \tan (c+d x)}{2 d}+\frac{\left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.892656, size = 235, normalized size = 3.56 \[ \frac{1}{32} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{12 \sin (d x)}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{14 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{14 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+4 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(4*x - (14*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (14*Log[
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + 1/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - 1/(d*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2])^2) + (12*Sin[d*x])/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - S
in[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/32

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Maple [A]  time = 0.03, size = 71, normalized size = 1.1 \begin{align*}{a}^{3}x+{\frac{{a}^{3}c}{d}}+{\frac{7\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x+1/d*a^3*c+7/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*tan(d*x+c)/d+1/2*a^3*sec(d*x+c)*tan(d*x+c)/d

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Maxima [A]  time = 1.14688, size = 123, normalized size = 1.86 \begin{align*} a^{3} x - \frac{a^{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^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac{3 \, a^{3} \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.75337, size = 251, normalized size = 3.8 \begin{align*} \frac{4 \, a^{3} d x \cos \left (d x + c\right )^{2} + 7 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 7 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{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+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(4*a^3*d*x*cos(d*x + c)^2 + 7*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 7*a^3*cos(d*x + c)^2*log(-sin(d*x
+ c) + 1) + 2*(6*a^3*cos(d*x + c) + a^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*} a^{3} \left (\int 1\, dx + \int 3 \sec{\left (c + d x \right )}\, dx + \int 3 \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(1, x) + Integral(3*sec(c + d*x), x) + Integral(3*sec(c + d*x)**2, x) + Integral(sec(c + d*x)**3
, x))

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Giac [A]  time = 1.35293, size = 135, normalized size = 2.05 \begin{align*} \frac{2 \,{\left (d x + c\right )} a^{3} + 7 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 7 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{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+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(d*x + c)*a^3 + 7*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 7*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2
*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d