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

Optimal. Leaf size=42 \[ \frac {a A \tan (c+d x)}{d}+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \sin (c+d x)}{d}+a C x \]

[Out]

a*C*x+a*A*arctanh(sin(d*x+c))/d+a*C*sin(d*x+c)/d+a*A*tan(d*x+c)/d

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Rubi [A]  time = 0.10, antiderivative size = 42, normalized size of antiderivative = 1.00, 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, 3023, 2735, 3770} \[ \frac {a A \tan (c+d x)}{d}+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \sin (c+d x)}{d}+a C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 54, normalized size = 1.29 \[ \frac {a A \tan (c+d x)}{d}+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \sin (c) \cos (d x)}{d}+\frac {a C \cos (c) \sin (d x)}{d}+a C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.15, size = 86, normalized size = 2.05 \[ \frac {2 \, C a d x \cos \left (d x + c\right ) + A a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - A a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a \cos \left (d x + c\right ) + A a\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.20, size = 117, normalized size = 2.79 \[ \frac {{\left (d x + c\right )} C a + A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

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)^2,x, algorithm="giac")

[Out]

((d*x + c)*C*a + A*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - A*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(A*a*tan(
1/2*d*x + 1/2*c)^3 - C*a*tan(1/2*d*x + 1/2*c)^3 + A*a*tan(1/2*d*x + 1/2*c) + C*a*tan(1/2*d*x + 1/2*c))/(tan(1/
2*d*x + 1/2*c)^4 - 1))/d

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maple [A]  time = 0.25, size = 57, normalized size = 1.36 \[ a C x +\frac {a A \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a C \sin \left (d x +c \right )}{d}+\frac {a C c}{d} \]

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)^2,x)

[Out]

a*C*x+a*A*tan(d*x+c)/d+1/d*a*A*ln(sec(d*x+c)+tan(d*x+c))+a*C*sin(d*x+c)/d+1/d*a*C*c

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maxima [A]  time = 0.63, size = 59, normalized size = 1.40 \[ \frac {2 \, {\left (d x + c\right )} C a + A a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a \sin \left (d x + c\right ) + 2 \, A a \tan \left (d x + c\right )}{2 \, d} \]

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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.88, size = 91, normalized size = 2.17 \[ \frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*a*sin(c + d*x))/d + (2*A*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a*atan(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2)))/d + (A*a*sin(c + d*x))/(d*cos(c + d*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

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)**2,x)

[Out]

a*(Integral(A*sec(c + d*x)**2, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*s
ec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**2, x))

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