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

Optimal. Leaf size=67 \[ \frac {a \left (a^2-b^2\right ) \sin (c+d x)}{d}+3 a^2 b x+\frac {3 a b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \sin (c+d x) (a+b \sec (c+d x))}{d} \]

[Out]

3*a^2*b*x+3*a*b^2*arctanh(sin(d*x+c))/d+a*(a^2-b^2)*sin(d*x+c)/d+b^2*(a+b*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.11, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3842, 4047, 8, 4045, 3770} \[ \frac {a \left (a^2-b^2\right ) \sin (c+d x)}{d}+3 a^2 b x+\frac {3 a b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \sin (c+d x) (a+b \sec (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a^2*b*x + (3*a*b^2*ArcTanh[Sin[c + d*x]])/d + (a*(a^2 - b^2)*Sin[c + d*x])/d + (b^2*(a + b*Sec[c + d*x])*Sin
[c + d*x])/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 3842

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(d*(m + n - 1)), In
t[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2) +
3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Integ
erQ[m])

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

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Mathematica [A]  time = 0.34, size = 88, normalized size = 1.31 \[ \frac {a^3 \sin (c+d x)+3 a b \left (a c+a d x-b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+b^3 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b*(a*c + a*d*x - b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])
+ a^3*Sin[c + d*x] + b^3*Tan[c + d*x])/d

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fricas [A]  time = 0.49, size = 94, normalized size = 1.40 \[ \frac {6 \, a^{2} b d x \cos \left (d x + c\right ) + 3 \, a b^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right ) + b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 131, normalized size = 1.96 \[ \frac {3 \, {\left (d x + c\right )} a^{2} b + 3 \, a b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.64, size = 68, normalized size = 1.01 \[ 3 a^{2} b x +\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} \sin \left (d x +c \right )}{d}+\frac {b^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 66, normalized size = 0.99 \[ \frac {6 \, {\left (d x + c\right )} a^{2} b + 3 \, a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{3} \sin \left (d x + c\right ) + 2 \, b^{3} \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.92, size = 97, normalized size = 1.45 \[ \frac {a^3\,\sin \left (c+d\,x\right )}{d}+\frac {b^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {6\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,a\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*sin(c + d*x))/d + (b^3*sin(c + d*x))/(d*cos(c + d*x)) + (6*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/
2)))/d + (6*a*b^2*atanh(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 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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