3.6 \(\int (a+a \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=58 \[ \frac{a (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A \tan (c+d x)}{d}+\frac{a A \tan (c+d x) \sec (c+d x)}{2 d}+a C x \]

[Out]

a*C*x + (a*(A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*A*Tan[c + d*x])/d + (a*A*Sec[c + d*x]*Tan[c + d*x])/(2*
d)

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Rubi [A]  time = 0.124825, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3032, 3021, 2735, 3770} \[ \frac{a (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A \tan (c+d x)}{d}+\frac{a A \tan (c+d x) \sec (c+d x)}{2 d}+a C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*C*x + (a*(A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*A*Tan[c + d*x])/d + (a*A*Sec[c + d*x]*Tan[c + d*x])/(2*
d)

Rule 3032

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1
))/(b^2*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b
*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f
*x] + b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

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

Mathematica [A]  time = 0.0301834, size = 67, normalized size = 1.16 \[ \frac{a A \tan (c+d x)}{d}+\frac{a A \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a C \tanh ^{-1}(\sin (c+d x))}{d}+a C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*C*x + (a*A*ArcTanh[Sin[c + d*x]])/(2*d) + (a*C*ArcTanh[Sin[c + d*x]])/d + (a*A*Tan[c + d*x])/d + (a*A*Sec[c
+ d*x]*Tan[c + d*x])/(2*d)

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Maple [A]  time = 0.089, size = 85, normalized size = 1.5 \begin{align*}{\frac{aA\tan \left ( dx+c \right ) }{d}}+aCx+{\frac{Cac}{d}}+{\frac{aA\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

[Out]

a*A*tan(d*x+c)/d+a*C*x+1/d*a*C*c+1/2*a*A*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a*A*ln(sec(d*x+c)+tan(d*x+c))+1/d*a*C*l
n(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.26223, size = 128, normalized size = 2.21 \begin{align*} \frac{4 \,{\left (d x + c\right )} C a - A a{\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 )} + 2 \, C a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/4*(4*(d*x + c)*C*a - A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) + 2*C*a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a*tan(d*x + c))/d

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Fricas [A]  time = 1.36035, size = 267, normalized size = 4.6 \begin{align*} \frac{4 \, C a d x \cos \left (d x + c\right )^{2} +{\left (A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a \cos \left (d x + c\right ) + A a\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+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(4*C*a*d*x*cos(d*x + c)^2 + (A + 2*C)*a*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (A + 2*C)*a*cos(d*x + c)^2*
log(-sin(d*x + c) + 1) + 2*(2*A*a*cos(d*x + c) + A*a)*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+a*cos(d*x+c))*(A+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.19806, size = 142, normalized size = 2.45 \begin{align*} \frac{2 \,{\left (d x + c\right )} C a +{\left (A a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (A a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, A a \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+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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