3.1.25 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \, dx\) [25]

Optimal. Leaf size=59 \[ \frac {7 a^3 x}{2}+\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

7/2*a^3*x+a^3*arctanh(sin(d*x+c))/d+3*a^3*sin(d*x+c)/d+1/2*a^3*cos(d*x+c)*sin(d*x+c)/d

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Rubi [A]
time = 0.05, antiderivative size = 59, 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 = {3876, 2717, 2715, 8, 3855} \begin {gather*} \frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {7 a^3 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2717

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

Rule 3855

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

Rule 3876

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 81, normalized size = 1.37 \begin {gather*} \frac {a^3 \left (14 d x-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 \sin (c+d x)+\sin (2 (c+d x))\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(14*d*x - 4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 12*Si
n[c + d*x] + Sin[2*(c + d*x)]))/(4*d)

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Maple [A]
time = 0.08, size = 71, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \left (d x +c \right )+3 a^{3} \sin \left (d x +c \right )+a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(71\)
default	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \left (d x +c \right )+3 a^{3} \sin \left (d x +c \right )+a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(71\)
risch	\(\frac {7 a^{3} x}{2}-\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d}\)	\(102\)
norman	\(\frac {\frac {7 a^{3} x}{2}+\frac {7 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {9 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-7 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 74, normalized size = 1.25 \begin {gather*} \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 12 \, {\left (d x + c\right )} a^{3} + 2 \, a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} \sin \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 + 12*(d*x + c)*a^3 + 2*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c)
- 1)) + 12*a^3*sin(d*x + c))/d

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Fricas [A]
time = 3.06, size = 65, normalized size = 1.10 \begin {gather*} \frac {7 \, a^{3} d x + a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} \cos \left (d x + c\right ) + 6 \, a^{3}\right )} \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(7*a^3*d*x + a^3*log(sin(d*x + c) + 1) - a^3*log(-sin(d*x + c) + 1) + (a^3*cos(d*x + c) + 6*a^3)*sin(d*x +
c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.48, size = 100, normalized size = 1.69 \begin {gather*} \frac {7 \, {\left (d x + c\right )} a^{3} + 2 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.73, size = 88, normalized size = 1.49 \begin {gather*} \frac {7\,a^3\,x}{2}+\frac {2\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+7\,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+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*a^3*x)/2 + (2*a^3*atanh(tan(c/2 + (d*x)/2)))/d + (5*a^3*tan(c/2 + (d*x)/2)^3 + 7*a^3*tan(c/2 + (d*x)/2))/(d
*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1))

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