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

Optimal. Leaf size=79 \[ \frac {1}{2} a x \left (a^2+6 b^2\right )+\frac {5 a^2 b \sin (c+d x)}{2 d}+\frac {a^2 \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))}{2 d}+\frac {b^3 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

1/2*a*(a^2+6*b^2)*x+b^3*arctanh(sin(d*x+c))/d+5/2*a^2*b*sin(d*x+c)/d+1/2*a^2*cos(d*x+c)*(a+b*sec(d*x+c))*sin(d
*x+c)/d

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Rubi [A]  time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3841, 4047, 8, 4045, 3770} \[ \frac {1}{2} a x \left (a^2+6 b^2\right )+\frac {5 a^2 b \sin (c+d x)}{2 d}+\frac {a^2 \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))}{2 d}+\frac {b^3 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 105, normalized size = 1.33 \[ \frac {a^3 \sin (2 (c+d x))+2 a \left (a^2+6 b^2\right ) (c+d x)+12 a^2 b \sin (c+d x)-4 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 72, normalized size = 0.91 \[ \frac {b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} + 6 \, a b^{2}\right )} d x + {\left (a^{3} \cos \left (d x + c\right ) + 6 \, a^{2} b\right )} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 137, normalized size = 1.73 \[ \frac {2 \, b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (a^{3} + 6 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.50, size = 90, normalized size = 1.14 \[ \frac {a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{3} x}{2}+\frac {a^{3} c}{2 d}+\frac {3 a^{2} b \sin \left (d x +c \right )}{d}+3 b^{2} a x +\frac {3 a \,b^{2} c}{d}+\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 76, normalized size = 0.96 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 12 \, {\left (d x + c\right )} a b^{2} + 2 \, b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} b \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.03, size = 123, normalized size = 1.56 \[ \frac {a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,a^2\,b\,\sin \left (c+d\,x\right )}{d}+\frac {6\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(a^3*sin(2*c + 2*d*x))/(4*d) + (3*a^2*b*sin(c + d*x))/d + (6*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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