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

Optimal. Leaf size=68 \[ -\frac{b \left (a^2-b^2\right ) \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \tan (c+d x) (a+b \cos (c+d x))}{d}+3 a b^2 x \]

[Out]

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

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Rubi [A]  time = 0.122414, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2792, 3023, 2735, 3770} \[ -\frac{b \left (a^2-b^2\right ) \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \tan (c+d x) (a+b \cos (c+d x))}{d}+3 a b^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.333023, size = 88, normalized size = 1.29 \[ \frac{a^3 \tan (c+d x)+3 a b \left (-a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+b c+b d x\right )+b^3 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.059, size = 68, normalized size = 1. \begin{align*} 3\,a{b}^{2}x+{\frac{{b}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2}c}{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)^2,x)

[Out]

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

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Maxima [A]  time = 0.983287, size = 89, normalized size = 1.31 \begin{align*} \frac{6 \,{\left (d x + c\right )} a b^{2} + 3 \, a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, b^{3} \sin \left (d x + c\right ) + 2 \, a^{3} \tan \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.0692, size = 246, normalized size = 3.62 \begin{align*} \frac{6 \, a b^{2} d x \cos \left (d x + c\right ) + 3 \, a^{2} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (b^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.42718, size = 174, normalized size = 2.56 \begin{align*} \frac{3 \,{\left (d x + c\right )} a b^{2} + 3 \, a^{2} b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} b \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} \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)^2,x, algorithm="giac")

[Out]

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