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

Optimal. Leaf size=154 \[ \frac {3 (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 b^2 d} \]

[Out]

3/7*(7*A+4*C)*hypergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/
2)+3/4*B*hypergeom([-2/3, 1/2],[1/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(4/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)+
3/7*C*(b*sec(d*x+c))^(4/3)*tan(d*x+c)/b^2/d

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Rubi [A]  time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac {3 (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(7*A + 4*C)*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(7*b*d*S
qrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(4/3)*Sin[c + d
*x])/(4*b^2*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(4/3)*Tan[c + d*x])/(7*b^2*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 \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx &=\frac {\int (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac {\int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{7/3} \, dx}{b^3}\\ &=\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d}+\frac {(7 A+4 C) \int (b \sec (c+d x))^{4/3} \, dx}{7 b^2}+\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 )^{7/3}} \, dx}{b^3}\\ &=\frac {3 B \, _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)}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d}+\frac {\left ((7 A+4 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 )^{4/3}} \, dx}{7 b^2}\\ &=\frac {3 (7 A+4 C) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \, _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)}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d}\\ \end {align*}

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Mathematica [C]  time = 3.24, size = 304, normalized size = 1.97 \[ \frac {3 b e^{-i c} \left (-1+e^{2 i c}\right ) \csc (c) \left (2 (7 A+4 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{7/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-e^{2 i (c+d x)}\right )-28 A e^{i (c+d x)}-56 A e^{3 i (c+d x)}-28 A e^{5 i (c+d x)}-7 B \left (1+e^{2 i (c+d x)}\right )^{7/3} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (c+d x)}\right )-7 B e^{4 i (c+d x)}+7 B-8 C e^{i (c+d x)}-40 C e^{3 i (c+d x)}-16 C e^{5 i (c+d x)}\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{28 d \left (1+e^{2 i (c+d x)}\right )^2 (b \sec (c+d x))^{5/3} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(-1 + E^((2*I)*c))*Csc[c]*(7*B - 28*A*E^(I*(c + d*x)) - 8*C*E^(I*(c + d*x)) - 56*A*E^((3*I)*(c + d*x)) -
40*C*E^((3*I)*(c + d*x)) - 7*B*E^((4*I)*(c + d*x)) - 28*A*E^((5*I)*(c + d*x)) - 16*C*E^((5*I)*(c + d*x)) - 7*B
*(1 + E^((2*I)*(c + d*x)))^(7/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(c + d*x))] + 2*(7*A + 4*C)*E^(I*(
c + d*x))*(1 + E^((2*I)*(c + d*x)))^(7/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -E^((2*I)*(c + d*x))])*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2))/(28*d*E^(I*c)*(1 + E^((2*I)*(c + d*x)))^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(
c + d*x)])*(b*Sec[c + d*x])^(5/3))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{3} + B \sec \left (d x + c\right )^{2} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}}{b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.76, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{2}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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