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

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

[Out]

b*C*x+1/2*a*(A+2*C)*arctanh(sin(d*x+c))/d+A*b*tan(d*x+c)/d+1/2*a*A*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.13, antiderivative size = 58, 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, 3021, 2735, 3770} \[ \frac {a (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {A b \tan (c+d x)}{d}+b C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 67, normalized size = 1.16 \[ \frac {a A \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A b \tan (c+d x)}{d}+b C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.68, size = 101, normalized size = 1.74 \[ \frac {4 \, C b d x \cos \left (d x + c\right )^{2} + {\left (A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A b \cos \left (d x + c\right ) + A a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*C*b*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*b*cos(d*x + c) + A*a)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [B]  time = 0.40, size = 132, normalized size = 2.28 \[ \frac {2 \, {\left (d x + c\right )} C b + {\left (A a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.25, size = 85, normalized size = 1.47 \[ \frac {a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A b \tan \left (d x +c \right )}{d}+b C x +\frac {C b c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.70, size = 95, normalized size = 1.64 \[ \frac {4 \, {\left (d x + c\right )} C b - A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A b \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*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*b - 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*b*tan(d*x + c))/d

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mupad [B]  time = 1.40, size = 135, normalized size = 2.33 \[ \frac {2\,C\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (C*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
2i)/d - (A*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d + (A*a*sin(c + d*x))/(2*d*cos(c + d*x)^2)
+ (A*b*sin(c + d*x))/(d*cos(c + d*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

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

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