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

Optimal. Leaf size=184 \[ \frac{16 a^2 (21 A+15 B+13 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (21 A+15 B+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac{2 (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

[Out]

(64*a^3*(21*A + 15*B + 13*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(21*A + 15*B + 13*C)*Sqr
t[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*a*(21*A + 15*B + 13*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x
])/(105*d) + (2*(9*B - 2*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*C*(a + a*Sec[c + d*x])^(7/2)*
Tan[c + d*x])/(9*a*d)

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Rubi [A]  time = 0.3447, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4082, 4001, 3793, 3792} \[ \frac{16 a^2 (21 A+15 B+13 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (21 A+15 B+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac{2 (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(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*(21*A + 15*B + 13*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(21*A + 15*B + 13*C)*Sqr
t[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*a*(21*A + 15*B + 13*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x
])/(105*d) + (2*(9*B - 2*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*C*(a + a*Sec[c + d*x])^(7/2)*
Tan[c + d*x])/(9*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))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{2 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (9 A+7 C)+\frac{1}{2} a (9 B-2 C) \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{1}{21} (21 A+15 B+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac{2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{1}{105} (8 a (21 A+15 B+13 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 (21 A+15 B+13 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{1}{315} \left (32 a^2 (21 A+15 B+13 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (21 A+15 B+13 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}\\ \end{align*}

Mathematica [A]  time = 2.44974, size = 156, normalized size = 0.85 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt{a (\sec (c+d x)+1)} (2 (882 A+1215 B+1396 C) \cos (c+d x)+4 (966 A+870 B+803 C) \cos (2 (c+d x))+588 A \cos (3 (c+d x))+903 A \cos (4 (c+d x))+2961 A+690 B \cos (3 (c+d x))+690 B \cos (4 (c+d x))+2790 B+584 C \cos (3 (c+d x))+584 C \cos (4 (c+d x))+2908 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(2961*A + 2790*B + 2908*C + 2*(882*A + 1215*B + 1396*C)*Cos[c + d*x] + 4*(966*A + 870*B + 803*C)*Cos[2*(c
+ d*x)] + 588*A*Cos[3*(c + d*x)] + 690*B*Cos[3*(c + d*x)] + 584*C*Cos[3*(c + d*x)] + 903*A*Cos[4*(c + d*x)] +
690*B*Cos[4*(c + d*x)] + 584*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/
(1260*d)

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Maple [A]  time = 0.29, size = 174, normalized size = 1. \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 903\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+690\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+584\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+294\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+345\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+292\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+63\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+180\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+219\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+45\,B\cos \left ( dx+c \right ) +130\,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)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/315/d*a^2*(-1+cos(d*x+c))*(903*A*cos(d*x+c)^4+690*B*cos(d*x+c)^4+584*C*cos(d*x+c)^4+294*A*cos(d*x+c)^3+345*
B*cos(d*x+c)^3+292*C*cos(d*x+c)^3+63*A*cos(d*x+c)^2+180*B*cos(d*x+c)^2+219*C*cos(d*x+c)^2+45*B*cos(d*x+c)+130*
C*cos(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)*(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

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Fricas [A]  time = 0.533747, size = 374, normalized size = 2.03 \begin{align*} \frac{2 \,{\left ({\left (903 \, A + 690 \, B + 584 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} +{\left (294 \, A + 345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A + 60 \, B + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, B + 26 \, C\right )} a^{2} \cos \left (d x + c\right ) + 35 \, 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 )}{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)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*((903*A + 690*B + 584*C)*a^2*cos(d*x + c)^4 + (294*A + 345*B + 292*C)*a^2*cos(d*x + c)^3 + 3*(21*A + 60*
B + 73*C)*a^2*cos(d*x + c)^2 + 5*(9*B + 26*C)*a^2*cos(d*x + c) + 35*C*a^2)*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)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 5.36341, size = 470, normalized size = 2.55 \begin{align*} \frac{8 \,{\left (315 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (1050 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 840 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 630 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (1323 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 945 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 819 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (189 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 135 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 117 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (21 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, \sqrt{2} C a^{7} \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)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

8/315*(315*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 315*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^7*sgn(cos(d
*x + c)) - (1050*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 840*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 630*sqrt(2)*C*a^7*sgn
(cos(d*x + c)) - (1323*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 945*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 819*sqrt(2)*C*a
^7*sgn(cos(d*x + c)) - 4*(189*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 135*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 117*sqrt
(2)*C*a^7*sgn(cos(d*x + c)) - 2*(21*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 15*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 13*
sqrt(2)*C*a^7*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)