3.485 \(\int \sec (c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=104 \[ \frac{2 a (15 A+5 B+7 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (5 B-2 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

[Out]

(2*a*(15*A + 5*B + 7*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(5*B - 2*C)*Sqrt[a + a*Sec[c + d*x]
]*Tan[c + d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d)

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Rubi [A]  time = 0.209287, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {4082, 4001, 3792} \[ \frac{2 a (15 A+5 B+7 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (5 B-2 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(15*A + 5*B + 7*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(5*B - 2*C)*Sqrt[a + a*Sec[c + d*x]
]*Tan[c + d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d)

Rule 4082

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

Rule 4001

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

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac{2 \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (5 A+3 C)+\frac{1}{2} a (5 B-2 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac{2 (5 B-2 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac{1}{15} (15 A+5 B+7 C) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (15 A+5 B+7 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-2 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.786311, size = 83, normalized size = 0.8 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt{a (\sec (c+d x)+1)} ((15 A+10 B+8 C) \cos (2 (c+d x))+15 A+2 (5 B+4 C) \cos (c+d x)+10 B+14 C)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*A + 10*B + 14*C + 2*(5*B + 4*C)*Cos[c + d*x] + (15*A + 10*B + 8*C)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Sqrt[
a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d)

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Maple [A]  time = 0.319, size = 105, normalized size = 1. \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 15\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+10\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5\,B\cos \left ( dx+c \right ) +4\,C\cos \left ( dx+c \right ) +3\,C \right ) }{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^2+10*B*cos(d*x+c)^2+8*C*cos(d*x+c)^2+5*B*cos(d*x+c)+4*C*cos(d*x+c)+3*
C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)

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Maxima [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(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.494679, size = 225, normalized size = 2.16 \begin{align*} \frac{2 \,{\left ({\left (15 \, A + 10 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*((15*A + 10*B + 8*C)*cos(d*x + c)^2 + (5*B + 4*C)*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
 c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x), x)

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Giac [B]  time = 4.54192, size = 302, normalized size = 2.9 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{2} A a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} B a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} C a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (30 \, \sqrt{2} A a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 20 \, \sqrt{2} B a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 10 \, \sqrt{2} C a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (15 \, \sqrt{2} A a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, \sqrt{2} B a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, \sqrt{2} C a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{15 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/15*(15*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 15*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^3*sgn(cos(d*x +
 c)) - (30*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 20*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 10*sqrt(2)*C*a^3*sgn(cos(d*x
 + c)) - (15*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 5*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 7*sqrt(2)*C*a^3*sgn(cos(d*x
 + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2
*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)