3.45 \(\int \frac{(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.169136, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{2 (A-B) \tan (c+d x)}{a d}+\frac{(3 A-2 B) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(3 A-2 B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

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

Mathematica [B]  time = 3.0809, size = 289, normalized size = 2.7 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (4 (B-A) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+\cos \left (\frac{1}{2} (c+d x)\right ) \left (-\frac{4 (A-B) \sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+(4 B-6 A) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+\frac{A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+6 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{2 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.115, size = 252, normalized size = 2.4 \begin{align*} -{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3\,A}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{B}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{3\,A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,A}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{B}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+cos(d*x+c)*a),x)

[Out]

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

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Maxima [B]  time = 1.02482, size = 381, normalized size = 3.56 \begin{align*} -\frac{A{\left (\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (d x + c\right )}{{\left (a - \frac{a \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 )}} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, 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)^3/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.37279, size = 385, normalized size = 3.6 \begin{align*} \frac{{\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (A - B\right )} \cos \left (d x + c\right )^{2} +{\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - A\right )} \sin \left (d x + c\right )}{4 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\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)^3/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{3}{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c)),x)

[Out]

(Integral(A*sec(c + d*x)**3/(cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)**3/(cos(c + d*x) + 1
), x))/a

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Giac [A]  time = 1.25401, size = 212, normalized size = 1.98 \begin{align*} \frac{\frac{{\left (3 \, A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{{\left (3 \, A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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} a}}{2 \, 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)^3/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

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