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

Optimal. Leaf size=147 \[ -\frac {3 (4 A+C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 b d} \]

[Out]

-3/8*(4*A+C)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)+3
*B*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*C*(b
*sec(d*x+c))^(1/3)*tan(d*x+c)/b/d

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Rubi [A]  time = 0.15, antiderivative size = 147, 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} \[ -\frac {3 (4 A+C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*A + C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(2/3)*Sqrt[
Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])
/(b*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*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 \frac {\sec (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 \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b}\\ &=\frac {\int \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b}+\frac {B \int (b \sec (c+d x))^{4/3} \, dx}{b^2}\\ &=\frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 b d}+\frac {(4 A+C) \int \sqrt [3]{b \sec (c+d x)} \, dx}{4 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 )^{4/3}} \, dx}{b^2}\\ &=\frac {3 B \, _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)}{b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 b d}+\frac {\left ((4 A+C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}} \, dx}{4 b}\\ &=\frac {3 B \, _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)}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 (4 A+C) \cos (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{8 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 b d}\\ \end {align*}

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Mathematica [C]  time = 1.94, size = 305, normalized size = 2.07 \[ \frac {3 \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sqrt [3]{\sec (c+d x)} (4 B \csc (c) \cos (d x)+C \tan (c+d x))-\frac {i \sqrt [3]{2} e^{-i (c+d x)} \sqrt [3]{\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\left (-1+e^{2 i c}\right ) (4 A+C) e^{i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (c+d x)}\right )+4 B \left (-1+e^{2 i c}\right ) \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-e^{2 i (c+d x)}\right )+4 B \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{2 b d \sec ^{\frac {7}{3}}(c+d x) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((-I)*2^(1/3)*(E^(I*(c + d*x))/(1 + E^((2*I
)*(c + d*x))))^(1/3)*(4*B*(1 + E^((2*I)*(c + d*x))) + 4*B*(-1 + E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^(1/3)*H
ypergeometric2F1[-1/3, 1/3, 2/3, -E^((2*I)*(c + d*x))] + (4*A + 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))]))/(E^(I*(c + d*x))*(-1 + E^((2*
I)*c))) + Sec[c + d*x]^(1/3)*(4*B*Cos[d*x]*Csc[c] + C*Tan[c + d*x])))/(2*b*d*(A + 2*C + 2*B*Cos[c + d*x] + A*C
os[2*(c + d*x)])*Sec[c + d*x]^(7/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x +c \right ) \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(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)*(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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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)*(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)*(b/cos(c + d*x))^(2/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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