3.53 \(\int \frac{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=107 \[ \frac{2 (5 A-2 B) \tan (c+d x)}{3 a^2 d}-\frac{(2 A-B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(2 A-B) \tan (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac{(A-B) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-(((2*A - B)*ArcTanh[Sin[c + d*x]])/(a^2*d)) + (2*(5*A - 2*B)*Tan[c + d*x])/(3*a^2*d) - ((2*A - B)*Tan[c + d*x
])/(a^2*d*(1 + Cos[c + d*x])) - ((A - B)*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.295281, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2978, 2748, 3767, 8, 3770} \[ \frac{2 (5 A-2 B) \tan (c+d x)}{3 a^2 d}-\frac{(2 A-B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(2 A-B) \tan (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac{(A-B) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*A - B)*ArcTanh[Sin[c + d*x]])/(a^2*d)) + (2*(5*A - 2*B)*Tan[c + d*x])/(3*a^2*d) - ((2*A - B)*Tan[c + d*x
])/(a^2*d*(1 + Cos[c + d*x])) - ((A - B)*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2978

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 \frac{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac{(A-B) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(a (4 A-B)-2 a (A-B) \cos (c+d x)) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(2 A-B) \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \left (2 a^2 (5 A-2 B)-3 a^2 (2 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{3 a^4}\\ &=-\frac{(2 A-B) \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(2 (5 A-2 B)) \int \sec ^2(c+d x) \, dx}{3 a^2}-\frac{(2 A-B) \int \sec (c+d x) \, dx}{a^2}\\ &=-\frac{(2 A-B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(2 A-B) \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(2 (5 A-2 B)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=-\frac{(2 A-B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{2 (5 A-2 B) \tan (c+d x)}{3 a^2 d}-\frac{(2 A-B) \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 1.51164, size = 264, normalized size = 2.47 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left ((A-B) \tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+(A-B) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+6 \cos ^3\left (\frac{1}{2} (c+d x)\right ) \left ((2 A-B) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\frac{A \sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}\right )+2 (7 A-4 B) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((A - B)*Sec[c/2]*Sin[(d*x)/2] + 2*(7*A - 4*B)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] +
6*Cos[(c + d*x)/2]^3*((2*A - B)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]]) + (A*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (A - B)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 0.095, size = 205, normalized size = 1.9 \begin{align*}{\frac{A}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{5\,A}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,B}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{{a}^{2}d}}-{\frac{B}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{{a}^{2}d}}+{\frac{B}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*A*tan(1/2*d*x+1/2*c)-3/2/d/a^2*B*t
an(1/2*d*x+1/2*c)+2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^2*A/(tan(1/2*d*x
+1/2*c)-1)-2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)+1/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^2*A/(tan(1/2*d*x+1/2*c)
+1)

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Maxima [B]  time = 1.04919, size = 329, normalized size = 3.07 \begin{align*} \frac{A{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*sin
(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) - B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^
3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c
) + 1) - 1)/a^2))/d

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Fricas [B]  time = 1.46332, size = 502, normalized size = 4.69 \begin{align*} -\frac{3 \,{\left ({\left (2 \, A - B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (2 \, A - B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (14 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(3*((2*A - B)*cos(d*x + c)^3 + 2*(2*A - B)*cos(d*x + c)^2 + (2*A - B)*cos(d*x + c))*log(sin(d*x + c) + 1)
- 3*((2*A - B)*cos(d*x + c)^3 + 2*(2*A - B)*cos(d*x + c)^2 + (2*A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) -
2*(2*(5*A - 2*B)*cos(d*x + c)^2 + (14*A - 5*B)*cos(d*x + c) + 3*A)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^
2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**2/(a+a*cos(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)*
*2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x))/a**2

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Giac [A]  time = 1.21405, size = 209, normalized size = 1.95 \begin{align*} -\frac{\frac{6 \,{\left (2 \, A - B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \,{\left (2 \, A - B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(6*(2*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(2*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2
+ 12*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2
*d*x + 1/2*c)^3 + 15*A*a^4*tan(1/2*d*x + 1/2*c) - 9*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d