3.501 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=229 \[ \frac{16 a^2 (165 A+143 B+125 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac{2 a (165 A+143 B+125 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac{2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d} \]

[Out]

(64*a^3*(165*A + 143*B + 125*C)*Tan[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(165*A + 143*B + 125
*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*(165*A + 143*B + 125*C)*(a + a*Sec[c + d*x])^(3/2)*
Tan[c + d*x])/(1155*d) + (2*(99*A - 22*B + 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*d) + (2*C*Sec[c
+ d*x]^2*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(11*d) + (2*(11*B + 5*C)*(a + a*Sec[c + d*x])^(7/2)*Tan[c +
d*x])/(99*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.586802, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{16 a^2 (165 A+143 B+125 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac{2 a (165 A+143 B+125 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac{2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(64*a^3*(165*A + 143*B + 125*C)*Tan[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(165*A + 143*B + 125
*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*(165*A + 143*B + 125*C)*(a + a*Sec[c + d*x])^(3/2)*
Tan[c + d*x])/(1155*d) + (2*(99*A - 22*B + 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*d) + (2*C*Sec[c
+ d*x]^2*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(11*d) + (2*(11*B + 5*C)*(a + a*Sec[c + d*x])^(7/2)*Tan[c +
d*x])/(99*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))^{5/2} \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))^{5/2} \tan (c+d x)}{11 d}+\frac{2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (11 A+4 C)+\frac{1}{2} a (11 B+5 C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{4 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{7}{4} a^2 (11 B+5 C)+\frac{1}{4} a^2 (99 A-22 B+26 C) \sec (c+d x)\right ) \, dx}{99 a^2}\\ &=\frac{2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{1}{231} (165 A+143 B+125 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac{2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{(8 a (165 A+143 B+125 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{1155}\\ &=\frac{16 a^2 (165 A+143 B+125 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{\left (32 a^2 (165 A+143 B+125 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx}{3465}\\ &=\frac{64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (165 A+143 B+125 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}\\ \end{align*}

Mathematica [A]  time = 1.66857, size = 188, normalized size = 0.82 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt{a (\sec (c+d x)+1)} ((49830 A+49654 B+50140 C) \cos (c+d x)+4 (4290 A+4642 B+4615 C) \cos (2 (c+d x))+22935 A \cos (3 (c+d x))+3795 A \cos (4 (c+d x))+3795 A \cos (5 (c+d x))+13365 A+20878 B \cos (3 (c+d x))+3212 B \cos (4 (c+d x))+3212 B \cos (5 (c+d x))+15356 B+18460 C \cos (3 (c+d x))+2840 C \cos (4 (c+d x))+2840 C \cos (5 (c+d x))+18140 C)}{13860 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(13365*A + 15356*B + 18140*C + (49830*A + 49654*B + 50140*C)*Cos[c + d*x] + 4*(4290*A + 4642*B + 4615*C)*
Cos[2*(c + d*x)] + 22935*A*Cos[3*(c + d*x)] + 20878*B*Cos[3*(c + d*x)] + 18460*C*Cos[3*(c + d*x)] + 3795*A*Cos
[4*(c + d*x)] + 3212*B*Cos[4*(c + d*x)] + 2840*C*Cos[4*(c + d*x)] + 3795*A*Cos[5*(c + d*x)] + 3212*B*Cos[5*(c
+ d*x)] + 2840*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(13860*d)

________________________________________________________________________________________

Maple [A]  time = 0.33, size = 207, normalized size = 0.9 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 7590\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+6424\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5680\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+3795\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+3212\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2840\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1980\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2409\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2130\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+495\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1430\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1775\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+385\,B\cos \left ( dx+c \right ) +1120\,C\cos \left ( dx+c \right ) +315\,C \right ) }{3465\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465/d*a^2*(-1+cos(d*x+c))*(7590*A*cos(d*x+c)^5+6424*B*cos(d*x+c)^5+5680*C*cos(d*x+c)^5+3795*A*cos(d*x+c)^4
+3212*B*cos(d*x+c)^4+2840*C*cos(d*x+c)^4+1980*A*cos(d*x+c)^3+2409*B*cos(d*x+c)^3+2130*C*cos(d*x+c)^3+495*A*cos
(d*x+c)^2+1430*B*cos(d*x+c)^2+1775*C*cos(d*x+c)^2+385*B*cos(d*x+c)+1120*C*cos(d*x+c)+315*C)*(a*(cos(d*x+c)+1)/
cos(d*x+c))^(1/2)/cos(d*x+c)^5/sin(d*x+c)

________________________________________________________________________________________

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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 0.543928, size = 456, normalized size = 1.99 \begin{align*} \frac{2 \,{\left (2 \,{\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} +{\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \,{\left (660 \, A + 803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \,{\left (99 \, A + 286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \,{\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/3465*(2*(3795*A + 3212*B + 2840*C)*a^2*cos(d*x + c)^5 + (3795*A + 3212*B + 2840*C)*a^2*cos(d*x + c)^4 + 3*(6
60*A + 803*B + 710*C)*a^2*cos(d*x + c)^3 + 5*(99*A + 286*B + 355*C)*a^2*cos(d*x + c)^2 + 35*(11*B + 32*C)*a^2*
cos(d*x + c) + 315*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x +
c)^5)

________________________________________________________________________________________

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))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 5.5264, size = 554, normalized size = 2.42 \begin{align*} \text{result too large to display} \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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-8/3465*(3465*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 3465*sqrt(2)*B*a^8*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^8*sgn(
cos(d*x + c)) - (12705*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 10395*sqrt(2)*B*a^8*sgn(cos(d*x + c)) + 8085*sqrt(2)*
C*a^8*sgn(cos(d*x + c)) - (19635*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 15939*sqrt(2)*B*a^8*sgn(cos(d*x + c)) + 150
15*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - (16335*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 14157*sqrt(2)*B*a^8*sgn(cos(d*x
+ c)) + 12375*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - 4*(1815*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 1573*sqrt(2)*B*a^8*s
gn(cos(d*x + c)) + 1375*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - 2*(165*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 143*sqrt(2)
*B*a^8*sgn(cos(d*x + c)) + 125*sqrt(2)*C*a^8*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x
+ 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)