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

Optimal. Leaf size=142 \[ -\frac{3 b (A-2 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}}-\frac{3 B \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{1}{2},\frac{4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac{3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \]

[Out]

(-3*b*(A - 2*C)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3)*Sq
rt[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Sec[c + d*x]
)^(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^(2/3))

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Rubi [A]  time = 0.135153, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4047, 3772, 2643, 4046} \[ -\frac{3 b (A-2 C) \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}}-\frac{3 B \sin (c+d x) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{2 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac{3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(A - 2*C)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3)*Sq
rt[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Sec[c + d*x]
)^(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^(2/3))

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx &=\frac{B \int \sqrt [3]{b \sec (c+d x)} \, dx}{b}+\int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx\\ &=\frac{3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}+(A-2 C) \int \frac{1}{(b \sec (c+d x))^{2/3}} \, dx+\frac{\left (B \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}{b}\\ &=-\frac{3 B \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)}{2 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}+\left ((A-2 C) \sqrt [3]{\frac{\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \left (\frac{\cos (c+d x)}{b}\right )^{2/3} \, dx\\ &=-\frac{3 B \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)}{2 b d \sqrt{\sin ^2(c+d x)}}-\frac{3 (A-2 C) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{5 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\\ \end{align*}

Mathematica [C]  time = 1.83104, size = 173, normalized size = 1.22 \[ \frac{3 e^{-i d x} (\sin (d x)-i \cos (d x)) \sqrt [3]{b \sec (c+d x)} \left ((A-2 C) e^{i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{5}{3},-e^{2 i (c+d x)}\right )+4 B \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (c+d x)}\right )-2 A \cos (c+d x)+4 i C \sin (c+d x)+4 C \cos (c+d x)\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(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)*((-I)*Cos[d*x] + Sin[d*x])*(-2*A*Cos[c + d*x] + 4*C*Cos[c + d*x] + 4*B*(1 + E^((2*I)
*(c + d*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(c + d*x))] + (A - 2*C)*E^(I*(c + d*x))*(1 + E^(
(2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -E^((2*I)*(c + d*x))] + (4*I)*C*Sin[c + d*x]))/(4*b*d
*E^(I*d*x))

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Maple [F]  time = 0.131, size = 0, normalized size = 0. \begin{align*} \int{(A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2}) \left ( b\sec \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*sec(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec{\left (c + d x \right )}\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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