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

Optimal. Leaf size=73 \[ \frac{1}{2} b x \left (6 a^2+b^2\right )+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a b^2 \sin (c+d x)}{2 d}+\frac{b^2 \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.11458, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2793, 3023, 2735, 3770} \[ \frac{1}{2} b x \left (6 a^2+b^2\right )+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a b^2 \sin (c+d x)}{2 d}+\frac{b^2 \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2793

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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

Mathematica [A]  time = 0.139916, size = 105, normalized size = 1.44 \[ \frac{2 b \left (6 a^2+b^2\right ) (c+d x)-4 a^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 a b^2 \sin (c+d x)+b^3 \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 90, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{a}^{2}bx+3\,{\frac{{a}^{2}bc}{d}}+3\,{\frac{a{b}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}x}{2}}+{\frac{{b}^{3}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.965056, size = 93, normalized size = 1.27 \begin{align*} \frac{12 \,{\left (d x + c\right )} a^{2} b +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3} + 4 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, a b^{2} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.98894, size = 176, normalized size = 2.41 \begin{align*} \frac{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 (6 \, a^{2} b + b^{3}\right )} d x +{\left (b^{3} \cos \left (d x + c\right ) + 6 \, a b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.43241, size = 185, normalized size = 2.53 \begin{align*} \frac{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 ) +{\left (6 \, a^{2} b + b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, a b^{2} \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} + 6 \, a b^{2} \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 )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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