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

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

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2792, 3021, 2735, 3770} \[ \frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^2 b \tan (c+d x)}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))}{2 d}+b^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2792

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

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^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+b \cos (c+d x))^3 \sec ^3(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (5 a^2 b+a \left (a^2+6 b^2\right ) \cos (c+d x)+2 b^3 \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {5 a^2 b \tan (c+d x)}{2 d}+\frac {a^2 (a+b \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (a \left (a^2+6 b^2\right )+2 b^3 \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 x+\frac {5 a^2 b \tan (c+d x)}{2 d}+\frac {a^2 (a+b \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a \left (a^2+6 b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=b^3 x+\frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^2 b \tan (c+d x)}{2 d}+\frac {a^2 (a+b \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 55, normalized size = 0.70 \[ \frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))+a^2 \tan (c+d x) (a \sec (c+d x)+6 b)+2 b^3 d x}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 112, normalized size = 1.42 \[ \frac {4 \, b^{3} d x \cos \left (d x + c\right )^{2} + {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, a^{2} b \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*b^3*d*x*cos(d*x + c)^2 + (a^3 + 6*a*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (a^3 + 6*a*b^2)*cos(d*x
+ c)^2*log(-sin(d*x + c) + 1) + 2*(6*a^2*b*cos(d*x + c) + a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 0.70, size = 143, normalized size = 1.81 \[ \frac {2 \, {\left (d x + c\right )} b^{3} + {\left (a^{3} + 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (a^{3} + 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*b^3 + (a^3 + 6*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (a^3 + 6*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 - 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*tan(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 = 0.10, size = 95, normalized size = 1.20 \[ \frac {a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+b^{3} x +\frac {c \,b^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 101, normalized size = 1.28 \[ \frac {4 \, {\left (d x + c\right )} b^{3} - a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} b \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*b^3 - 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
)) + 6*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*a^2*b*tan(d*x + c))/d

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mupad [B]  time = 0.67, size = 136, normalized size = 1.72 \[ \frac {a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {6\,a\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,a^2\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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