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

Optimal. Leaf size=134 \[ -\frac{2 b^2 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2 B-2 a b C+2 b^2 B\right )}{2 a^3}-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \sin (c+d x) \cos (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.478919, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4034, 4104, 3919, 3831, 2659, 208} \[ -\frac{2 b^2 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2 B-2 a b C+2 b^2 B\right )}{2 a^3}-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \sin (c+d x) \cos (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4072

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

Rule 4034

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && 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 (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\int \frac{\cos ^2(c+d x) (B+C \sec (c+d x))}{a+b \sec (c+d x)} \, dx\\ &=\frac{B \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\int \frac{\cos (c+d x) \left (2 (b B-a C)-a B \sec (c+d x)-b B \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a}\\ &=-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 a d}+\frac{\int \frac{a^2 B+2 b^2 B-2 a b C+a b B \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2}\\ &=\frac{\left (a^2 B+2 b^2 B-2 a b C\right ) x}{2 a^3}-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\left (b^2 (b B-a C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3}\\ &=\frac{\left (a^2 B+2 b^2 B-2 a b C\right ) x}{2 a^3}-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{(b (b B-a C)) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^3}\\ &=\frac{\left (a^2 B+2 b^2 B-2 a b C\right ) x}{2 a^3}-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{(2 b (b B-a C)) \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^3 d}\\ &=\frac{\left (a^2 B+2 b^2 B-2 a b C\right ) x}{2 a^3}-\frac{2 b^2 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 \sqrt{a-b} \sqrt{a+b} d}-\frac{(b B-a C) \sin (c+d x)}{a^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 a d}\\ \end{align*}

Mathematica [A]  time = 0.32288, size = 121, normalized size = 0.9 \[ \frac{2 (c+d x) \left (a^2 B-2 a b C+2 b^2 B\right )+\frac{8 b^2 (b B-a C) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a^2 B \sin (2 (c+d x))+4 a (a C-b B) \sin (c+d x)}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2*B + 2*b^2*B - 2*a*b*C)*(c + d*x) + (8*b^2*(b*B - a*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b
^2]])/Sqrt[a^2 - b^2] + 4*a*(-(b*B) + a*C)*Sin[c + d*x] + a^2*B*Sin[2*(c + d*x)])/(4*a^3*d)

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Maple [B]  time = 0.116, size = 367, normalized size = 2.7 \begin{align*} -{\frac{B}{ad} \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 ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}Bb}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) Bb}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{B}{ad}\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{b}^{2}}{d{a}^{3}}}-2\,{\frac{C\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) b}{d{a}^{2}}}-2\,{\frac{B{b}^{3}}{d{a}^{3}\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 ) }+2\,{\frac{{b}^{2}C}{d{a}^{2}\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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^3*B*b+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*C+1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+
1/2*c)-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*B*b+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+
1/2*c)*C+1/a/d*arctan(tan(1/2*d*x+1/2*c))*B+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B*b^2-2/d/a^2*C*arctan(tan(1/2*
d*x+1/2*c))*b-2/d*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+2/d*b^2/
a^2/((a+b)*(a-b))^(1/2)*arctanh((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*(B*sec(d*x+c)+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 = 0.574448, size = 934, normalized size = 6.97 \begin{align*} \left [\frac{{\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} d x -{\left (C a b^{2} - B b^{3}\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 \, C a^{4} - 2 \, B a^{3} b - 2 \, C a^{2} b^{2} + 2 \, B a b^{3} +{\left (B a^{4} - B a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )} d}, \frac{{\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} d x + 2 \,{\left (C a b^{2} - B b^{3}\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 \, C a^{4} - 2 \, B a^{3} b - 2 \, C a^{2} b^{2} + 2 \, B a b^{3} +{\left (B a^{4} - B a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.22528, size = 306, normalized size = 2.28 \begin{align*} \frac{\frac{{\left (B a^{2} - 2 \, C a b + 2 \, B b^{2}\right )}{\left (d x + c\right )}}{a^{3}} + \frac{4 \,{\left (C a b^{2} - B b^{3}\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^{3}} - \frac{2 \,{\left (B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B 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}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((B*a^2 - 2*C*a*b + 2*B*b^2)*(d*x + c)/a^3 + 4*(C*a*b^2 - B*b^3)*(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^3)
- 2*(B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 - B*a*tan(1/2*d
*x + 1/2*c) - 2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d