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

Optimal. Leaf size=187 \[ \frac{8 a^2 (63 A+57 B+47 C) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{315 d}+\frac{2 a (63 A+57 B+47 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{2 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{21 a d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d} \]

[Out]

(8*a^2*(63*A + 57*B + 47*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(63*A + 57*B + 47*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*(63*A - 18*B + 22*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(31
5*d) + (2*C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(9*d) + (2*(3*B + C)*(a + a*Sec[c + d*x])^
(5/2)*Tan[c + d*x])/(21*a*d)

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Rubi [A]  time = 0.515844, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {4088, 4010, 4001, 3793, 3792} \[ \frac{8 a^2 (63 A+57 B+47 C) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{315 d}+\frac{2 a (63 A+57 B+47 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{2 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{21 a d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(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*(63*A + 57*B + 47*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(63*A + 57*B + 47*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*(63*A - 18*B + 22*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(31
5*d) + (2*C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(9*d) + (2*(3*B + C)*(a + a*Sec[c + d*x])^
(5/2)*Tan[c + d*x])/(21*a*d)

Rule 4088

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !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 ^2(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 \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac{2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (9 A+4 C)+\frac{3}{2} a (3 B+C) \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}+\frac{4 \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{15}{4} a^2 (3 B+C)+\frac{1}{4} a^2 (63 A-18 B+22 C) \sec (c+d x)\right ) \, dx}{63 a^2}\\ &=\frac{2 (63 A-18 B+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}+\frac{1}{105} (63 A+57 B+47 C) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{2 a (63 A+57 B+47 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 (63 A-18 B+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}+\frac{1}{315} (4 a (63 A+57 B+47 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{8 a^2 (63 A+57 B+47 C) \tan (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (63 A+57 B+47 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 (63 A-18 B+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}\\ \end{align*}

Mathematica [A]  time = 2.14844, size = 152, normalized size = 0.81 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt{a (\sec (c+d x)+1)} ((567 A+648 B+748 C) \cos (c+d x)+(882 A+858 B+748 C) \cos (2 (c+d x))+189 A \cos (3 (c+d x))+189 A \cos (4 (c+d x))+693 A+156 B \cos (3 (c+d x))+156 B \cos (4 (c+d x))+702 B+136 C \cos (3 (c+d x))+136 C \cos (4 (c+d x))+752 C)}{630 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(693*A + 702*B + 752*C + (567*A + 648*B + 748*C)*Cos[c + d*x] + (882*A + 858*B + 748*C)*Cos[2*(c + d*x)] +
189*A*Cos[3*(c + d*x)] + 156*B*Cos[3*(c + d*x)] + 136*C*Cos[3*(c + d*x)] + 189*A*Cos[4*(c + d*x)] + 156*B*Cos[
4*(c + d*x)] + 136*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(630*d)

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Maple [A]  time = 0.315, size = 172, normalized size = 0.9 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 378\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+312\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+272\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+189\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+156\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+136\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+63\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+117\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+102\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+45\,B\cos \left ( dx+c \right ) +85\,C\cos \left ( dx+c \right ) +35\,C \right ) }{315\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/315/d*a*(-1+cos(d*x+c))*(378*A*cos(d*x+c)^4+312*B*cos(d*x+c)^4+272*C*cos(d*x+c)^4+189*A*cos(d*x+c)^3+156*B*
cos(d*x+c)^3+136*C*cos(d*x+c)^3+63*A*cos(d*x+c)^2+117*B*cos(d*x+c)^2+102*C*cos(d*x+c)^2+45*B*cos(d*x+c)+85*C*c
os(d*x+c)+35*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^4/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)^2*(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 = 0.50904, size = 363, normalized size = 1.94 \begin{align*} \frac{2 \,{\left (2 \,{\left (189 \, A + 156 \, B + 136 \, C\right )} a \cos \left (d x + c\right )^{4} +{\left (189 \, A + 156 \, B + 136 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A + 39 \, B + 34 \, C\right )} a \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, B + 17 \, C\right )} a \cos \left (d x + c\right ) + 35 \, 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 )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(2*(189*A + 156*B + 136*C)*a*cos(d*x + c)^4 + (189*A + 156*B + 136*C)*a*cos(d*x + c)^3 + 3*(21*A + 39*B
+ 34*C)*a*cos(d*x + c)^2 + 5*(9*B + 17*C)*a*cos(d*x + c) + 35*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin
(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

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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)**2*(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 = 5.03251, size = 470, normalized size = 2.51 \begin{align*} \frac{4 \,{\left (315 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (945 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 735 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 525 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (1071 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 819 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 819 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (567 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 513 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 423 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (63 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 57 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, \sqrt{2} C a^{6} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{315 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{4} \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)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

4/315*(315*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 315*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^6*sgn(cos(d
*x + c)) - (945*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 735*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 525*sqrt(2)*C*a^6*sgn(
cos(d*x + c)) - (1071*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 819*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 819*sqrt(2)*C*a^
6*sgn(cos(d*x + c)) - (567*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 513*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 423*sqrt(2)
*C*a^6*sgn(cos(d*x + c)) - 2*(63*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 57*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 47*sqr
t(2)*C*a^6*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)