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 + (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)

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Rubi [A]  time = 0.134308, antiderivative size = 79, 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, 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 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 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 ^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*}

Mathematica [A]  time = 0.175937, size = 55, normalized size = 0.7 \[ \frac{a \left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))+a^2 \tan (c+d x) (a \sec (c+d x)+6 b)+2 b^3 d x}{2 d} \]

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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Maple [A]  time = 0.06, size = 95, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{2}b\tan \left ( dx+c \right ) }{d}}+3\,{\frac{a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{3}x+{\frac{{b}^{3}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)^3,x)

[Out]

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

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Maxima [A]  time = 0.965618, size = 136, normalized size = 1.72 \begin{align*} \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} \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)^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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Fricas [A]  time = 1.99616, size = 281, normalized size = 3.56 \begin{align*} \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}} \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)^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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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.80078, size = 193, normalized size = 2.44 \begin{align*} \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} \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)^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