3.495 \(\int \frac{\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=148 \[ \frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{b x \left (a^2+2 b^2\right )}{2 a^4}-\frac{b \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

[Out]

-(b*(a^2 + 2*b^2)*x)/(2*a^4) + (2*b^4*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqrt[a - b]*Sq
rt[a + b]*d) + ((2*a^2 + 3*b^2)*Sin[c + d*x])/(3*a^3*d) - (b*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) + (Cos[c + d
*x]^2*Sin[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.458556, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3853, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{b x \left (a^2+2 b^2\right )}{2 a^4}-\frac{b \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(a^2 + 2*b^2)*x)/(2*a^4) + (2*b^4*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqrt[a - b]*Sq
rt[a + b]*d) + ((2*a^2 + 3*b^2)*Sin[c + d*x])/(3*a^3*d) - (b*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) + (Cos[c + d
*x]^2*Sin[c + d*x])/(3*a*d)

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx &=\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\int \frac{\cos ^2(c+d x) \left (-3 b+2 a \sec (c+d x)+2 b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a}\\ &=-\frac{b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{\cos (c+d x) \left (-2 \left (2 a^2+3 b^2\right )-a b \sec (c+d x)+3 b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^2}\\ &=\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}-\frac{b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\int \frac{-3 b \left (a^2+2 b^2\right )-3 a b^2 \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac{b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}-\frac{b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{b^4 \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac{b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}-\frac{b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{b^3 \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac{b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}-\frac{b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=-\frac{b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac{2 b^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 \sqrt{a-b} \sqrt{a+b} d}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}-\frac{b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.31313, size = 122, normalized size = 0.82 \[ \frac{-6 b \left (a^2+2 b^2\right ) (c+d x)+3 a \left (3 a^2+4 b^2\right ) \sin (c+d x)-\frac{24 b^4 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-3 a^2 b \sin (2 (c+d x))+a^3 \sin (3 (c+d x))}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*b*(a^2 + 2*b^2)*(c + d*x) - (24*b^4*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2]
+ 3*a*(3*a^2 + 4*b^2)*Sin[c + d*x] - 3*a^2*b*Sin[2*(c + d*x)] + a^3*Sin[3*(c + d*x)])/(12*a^4*d)

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Maple [B]  time = 0.07, size = 367, normalized size = 2.5 \begin{align*} 2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{b}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{b}^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{4}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ){b}^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{b}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-{\frac{b}{d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){b}^{3}}{d{a}^{4}}}+2\,{\frac{{b}^{4}}{d{a}^{4}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*
b+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*b^2+4/3/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1
/2*c)^3+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*b^2+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d
*x+1/2*c)+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*b^2-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2
*d*x+1/2*c)*b-1/d/a^2*b*arctan(tan(1/2*d*x+1/2*c))-2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*b^3+2/d*b^4/a^4/((a+b)*(
a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87652, size = 875, normalized size = 5.91 \begin{align*} \left [\frac{3 \, \sqrt{a^{2} - b^{2}} b^{4} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 3 \,{\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} d x +{\left (4 \, a^{5} + 2 \, a^{3} b^{2} - 6 \, a b^{4} + 2 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}, \frac{6 \, \sqrt{-a^{2} + b^{2}} b^{4} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - 3 \,{\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} d x +{\left (4 \, a^{5} + 2 \, a^{3} b^{2} - 6 \, a b^{4} + 2 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(a^2 - b^2)*b^4*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(
d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 3*(a^4*b + a^2*b^
3 - 2*b^5)*d*x + (4*a^5 + 2*a^3*b^2 - 6*a*b^4 + 2*(a^5 - a^3*b^2)*cos(d*x + c)^2 - 3*(a^4*b - a^2*b^3)*cos(d*x
+ c))*sin(d*x + c))/((a^6 - a^4*b^2)*d), 1/6*(6*sqrt(-a^2 + b^2)*b^4*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c)
+ a)/((a^2 - b^2)*sin(d*x + c))) - 3*(a^4*b + a^2*b^3 - 2*b^5)*d*x + (4*a^5 + 2*a^3*b^2 - 6*a*b^4 + 2*(a^5 -
a^3*b^2)*cos(d*x + c)^2 - 3*(a^4*b - a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d)]

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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(cos(d*x+c)**3/(a+b*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.34216, size = 336, normalized size = 2.27 \begin{align*} \frac{\frac{12 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{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 )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b^{4}}{\sqrt{-a^{2} + b^{2}} a^{4}} - \frac{3 \,{\left (a^{2} b + 2 \, b^{3}\right )}{\left (d x + c\right )}}{a^{4}} + \frac{2 \,{\left (6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, b^{2} \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 )}^{3} a^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(12*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x +
1/2*c))/sqrt(-a^2 + b^2)))*b^4/(sqrt(-a^2 + b^2)*a^4) - 3*(a^2*b + 2*b^3)*(d*x + c)/a^4 + 2*(6*a^2*tan(1/2*d*x
+ 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*
b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2
*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d