3.28 \(\int (a+a \cos (c+d x))^3 \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=48 \[ \frac {a^3 \sin (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{d}+3 a^3 x \]

[Out]

3*a^3*x+3*a^3*arctanh(sin(d*x+c))/d+a^3*sin(d*x+c)/d+a^3*tan(d*x+c)/d

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Rubi [A]  time = 0.07, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2757, 2637, 3770, 3767, 8} \[ \frac {a^3 \sin (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{d}+3 a^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2,x]

[Out]

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

Rule 8

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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

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Mathematica [B]  time = 0.73, size = 211, normalized size = 4.40 \[ \frac {1}{8} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\sin (c) \cos (d x)}{d}+\frac {\cos (c) \sin (d x)}{d}+\frac {\sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+3 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(3*x - (3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (3*Log[Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (Cos[d*x]*Sin[c])/d + (Cos[c]*Sin[d*x])/d + Sin[(d*x)/2]/(d*(Cos[c/2]
- Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + Sin[(d*x)/2]/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]))))/8

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fricas [A]  time = 0.80, size = 91, normalized size = 1.90 \[ \frac {6 \, a^{3} d x \cos \left (d x + c\right ) + 3 \, a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.61, size = 80, normalized size = 1.67 \[ \frac {3 \, {\left (d x + c\right )} a^{3} + 3 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*sec(d*x+c)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 65, normalized size = 1.35 \[ 3 a^{3} x +\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} \sin \left (d x +c \right )}{d}+\frac {a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*sec(d*x+c)^2,x)

[Out]

3*a^3*x+3/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+a^3*sin(d*x+c)/d+a^3*tan(d*x+c)/d+3/d*a^3*c

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maxima [A]  time = 1.00, size = 64, normalized size = 1.33 \[ \frac {6 \, {\left (d x + c\right )} a^{3} + 3 \, a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3} \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.41, size = 57, normalized size = 1.19 \[ 3\,a^3\,x+\frac {6\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {4\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^3/cos(c + d*x)^2,x)

[Out]

3*a^3*x + (6*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (4*a^3*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^4 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int 3 \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**3*sec(d*x+c)**2,x)

[Out]

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

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