3.48 \(\int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=151 \[ \frac {3 (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 b d} \]

[Out]

3/40*(10*A+7*C)*hypergeom([-2/3, 1/2],[1/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(4/3)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/
2)+3/7*B*hypergeom([-7/6, 1/2],[-1/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(7/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3
/10*C*(b*sec(d*x+c))^(7/3)*tan(d*x+c)/b/d

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac {3 (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*(10*A + 7*C)*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(4/3)*Sin[c + d*x])/(40*d*S
qrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c +
d*x])/(7*b*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(7/3)*Tan[c + d*x])/(10*b*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 4046

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

Rule 4047

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

Rubi steps

\begin {align*} \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (b \sec (c+d x))^{7/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b}\\ &=\frac {\int (b \sec (c+d x))^{7/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b}+\frac {B \int (b \sec (c+d x))^{10/3} \, dx}{b^2}\\ &=\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac {(10 A+7 C) \int (b \sec (c+d x))^{7/3} \, dx}{10 b}+\frac {\left (B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{10/3}} \, dx}{b^2}\\ &=\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac {3 b B \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {\left ((10 A+7 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}} \, dx}{10 b}\\ &=\frac {3 (10 A+7 C) \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac {3 b B \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 7.02, size = 465, normalized size = 3.08 \[ \frac {\frac {\cos ^4(c+d x) (b \sec (c+d x))^{7/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {3 \sec (c) \sec (c+d x) (70 A \sin (d x)+40 B \sin (c)+49 C \sin (d x))}{140 d}+\frac {3 (10 A+7 C) \tan (c)}{20 d}+\frac {3 \sec (c) \sec ^2(c+d x) (10 B \sin (d x)+7 C \sin (c))}{35 d}+\frac {24 B \csc (c) \cos (d x)}{7 d}+\frac {3 C \sec (c) \sin (d x) \sec ^3(c+d x)}{5 d}\right )}{A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C}-\frac {3 i e^{-i (c+d x)} \sqrt [3]{\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} (b \sec (c+d x))^{7/3} \left (7 \left (-1+e^{2 i c}\right ) (10 A+7 C) e^{i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (c+d x)}\right )+160 B \left (-1+e^{2 i c}\right ) \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-e^{2 i (c+d x)}\right )+160 B \left (1+e^{2 i (c+d x)}\right )\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{70\ 2^{2/3} \left (-1+e^{2 i c}\right ) d \sec ^{\frac {13}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((((-3*I)/70)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1/3)*(160*B*(1 + E^((2*I)*(c + d*x))) + 160*B*(-1 +
E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[-1/3, 1/3, 2/3, -E^((2*I)*(c + d*x))] + 7*(10*
A + 7*C)*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -
E^((2*I)*(c + d*x))])*(b*Sec[c + d*x])^(7/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(2^(2/3)*d*E^(I*(c + d*x
))*(-1 + E^((2*I)*c))*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(13/3)) + (Cos[c + d*x]^4
*(b*Sec[c + d*x])^(7/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((24*B*Cos[d*x]*Csc[c])/(7*d) + (3*C*Sec[c]*Se
c[c + d*x]^3*Sin[d*x])/(5*d) + (3*Sec[c]*Sec[c + d*x]^2*(7*C*Sin[c] + 10*B*Sin[d*x]))/(35*d) + (3*Sec[c]*Sec[c
+ d*x]*(40*B*Sin[c] + 70*A*Sin[d*x] + 49*C*Sin[d*x]))/(140*d) + (3*(10*A + 7*C)*Tan[c])/(20*d)))/(A + 2*C + 2
*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))/b

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \sec \left (d x + c\right )^{4} + B b \sec \left (d x + c\right )^{3} + A b \sec \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b*sec(d*x + c)^4 + B*b*sec(d*x + c)^3 + A*b*sec(d*x + c)^2)*(b*sec(d*x + c))^(1/3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*sec(d*x + c), x)

________________________________________________________________________________________

maple [F]  time = 1.05, size = 0, normalized size = 0.00 \[ \int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x),x)

[Out]

int(((b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________