3.682 \(\int \frac{\cos ^3(c+d x) (A+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=175 \[ \frac{\left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{b x \left (a^2 (A+2 C)+2 A b^2\right )}{2 a^4}-\frac{A b \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.605461, antiderivative size = 173, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4105, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{b x \left (\frac{2 A b^2}{a^2}+A+2 C\right )}{2 a^2}-\frac{A b \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4105

Int[((A_.) + 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*(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, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]

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

Mathematica [A]  time = 0.500142, size = 149, normalized size = 0.85 \[ \frac{-6 b (c+d x) \left (a^2 (A+2 C)+2 A b^2\right )+3 a \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)-\frac{24 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-3 a^2 A b \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.118, size = 551, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(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*A+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^
5*A*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*A*b^2+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d
*x+1/2*c)^5*C+4/3/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan
(1/2*d*x+1/2*c)^3*A*b^2+4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*C+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)
^3*tan(1/2*d*x+1/2*c)*A+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b^2+2/d/a/(1+tan(1/2*d*x+1/2*c
)^2)^3*tan(1/2*d*x+1/2*c)*C-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b-1/d*A/a^2*b*arctan(tan(1
/2*d*x+1/2*c))-2/d/a^4*A*arctan(tan(1/2*d*x+1/2*c))*b^3-2/d/a^2*C*arctan(tan(1/2*d*x+1/2*c))*b+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))*A+2/d*b^2/a^2/((a+b)*(a-b))^(1/2)*arcta
nh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C

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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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.596147, size = 1067, normalized size = 6.1 \begin{align*} \left [-\frac{3 \,{\left ({\left (A + 2 \, C\right )} a^{4} b +{\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, A b^{5}\right )} d x - 3 \,{\left (C a^{2} b^{2} + A b^{4}\right )} \sqrt{a^{2} - b^{2}} \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 ) -{\left (2 \,{\left (2 \, A + 3 \, C\right )} a^{5} + 2 \,{\left (A - 3 \, C\right )} a^{3} b^{2} - 6 \, A a b^{4} + 2 \,{\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (A a^{4} b - A 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{3 \,{\left ({\left (A + 2 \, C\right )} a^{4} b +{\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, A b^{5}\right )} d x - 6 \,{\left (C a^{2} b^{2} + A b^{4}\right )} \sqrt{-a^{2} + b^{2}} \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 ) -{\left (2 \,{\left (2 \, A + 3 \, C\right )} a^{5} + 2 \,{\left (A - 3 \, C\right )} a^{3} b^{2} - 6 \, A a b^{4} + 2 \,{\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (A a^{4} b - A 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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.21354, size = 440, normalized size = 2.51 \begin{align*} -\frac{\frac{3 \,{\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )}{\left (d x + c\right )}}{a^{4}} - \frac{12 \,{\left (C a^{2} b^{2} + A b^{4}\right )}{\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 )}}{\sqrt{-a^{2} + b^{2}} a^{4}} - \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A 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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(A*a^2*b + 2*C*a^2*b + 2*A*b^3)*(d*x + c)/a^4 - 12*(C*a^2*b^2 + A*b^4)*(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)))/(sqrt(-a^2 +
b^2)*a^4) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)
^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*A*
b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x
+ 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d