3.493 \(\int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=144 \[ \frac{8 a^2 (35 A+21 B+19 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (35 A+21 B+19 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d} \]

[Out]

(8*a^2*(35*A + 21*B + 19*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(35*A + 21*B + 19*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*B - 2*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*
C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*a*d)

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Rubi [A]  time = 0.283473, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4082, 4001, 3793, 3792} \[ \frac{8 a^2 (35 A+21 B+19 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (35 A+21 B+19 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(35*A + 21*B + 19*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(35*A + 21*B + 19*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*B - 2*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*
C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*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 3793

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

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) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac{2 \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (7 A+5 C)+\frac{1}{2} a (7 B-2 C) \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 (7 B-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac{1}{35} (35 A+21 B+19 C) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{2 a (35 A+21 B+19 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 B-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac{1}{105} (4 a (35 A+21 B+19 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{8 a^2 (35 A+21 B+19 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (35 A+21 B+19 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 B-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}\\ \end{align*}

Mathematica [A]  time = 1.63287, size = 120, normalized size = 0.83 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt{a (\sec (c+d x)+1)} ((525 A+462 B+468 C) \cos (c+d x)+2 (35 A+63 B+52 C) \cos (2 (c+d x))+175 A \cos (3 (c+d x))+70 A+126 B \cos (3 (c+d x))+126 B+104 C \cos (3 (c+d x))+164 C)}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(70*A + 126*B + 164*C + (525*A + 462*B + 468*C)*Cos[c + d*x] + 2*(35*A + 63*B + 52*C)*Cos[2*(c + d*x)] + 17
5*A*Cos[3*(c + d*x)] + 126*B*Cos[3*(c + d*x)] + 104*C*Cos[3*(c + d*x)])*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x
])]*Tan[(c + d*x)/2])/(210*d)

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Maple [A]  time = 0.296, size = 139, normalized size = 1. \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 175\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+126\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+104\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+35\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+52\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,B\cos \left ( dx+c \right ) +39\,C\cos \left ( dx+c \right ) +15\,C \right ) }{105\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/105/d*a*(-1+cos(d*x+c))*(175*A*cos(d*x+c)^3+126*B*cos(d*x+c)^3+104*C*cos(d*x+c)^3+35*A*cos(d*x+c)^2+63*B*co
s(d*x+c)^2+52*C*cos(d*x+c)^2+21*B*cos(d*x+c)+39*C*cos(d*x+c)+15*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x
+c)^3/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+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.501427, size = 300, normalized size = 2.08 \begin{align*} \frac{2 \,{\left ({\left (175 \, A + 126 \, B + 104 \, C\right )} a \cos \left (d x + c\right )^{3} +{\left (35 \, A + 63 \, B + 52 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, B + 13 \, C\right )} a \cos \left (d x + c\right ) + 15 \, C a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*((175*A + 126*B + 104*C)*a*cos(d*x + c)^3 + (35*A + 63*B + 52*C)*a*cos(d*x + c)^2 + 3*(7*B + 13*C)*a*cos
(d*x + c) + 15*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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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(sec(d*x+c)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 4.89244, size = 386, normalized size = 2.68 \begin{align*} -\frac{4 \,{\left (105 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} B a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (280 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 210 \, \sqrt{2} B a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 140 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (245 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt{2} B a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 133 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (35 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 21 \, \sqrt{2} B a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 19 \, \sqrt{2} C a^{5} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{105 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \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+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-4/105*(105*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^5*sgn(cos(
d*x + c)) - (280*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 210*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 140*sqrt(2)*C*a^5*sgn
(cos(d*x + c)) - (245*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 147*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 133*sqrt(2)*C*a^
5*sgn(cos(d*x + c)) - 2*(35*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 21*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 19*sqrt(2)*
C*a^5*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)^2)*tan(1/2*d*x +
1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)