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

Optimal. Leaf size=146 \[ -\frac{3 (A+4 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{2},\frac{7}{6},\cos ^2(c+d x)\right )}{4 b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}+\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}} \]

[Out]

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

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Rubi [A]  time = 0.138117, antiderivative size = 146, 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, 4045} \[ -\frac{3 (A+4 C) \sin (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right )}{4 b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}-\frac{3 B \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, 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))^{4/3}} \, dx &=\frac{B \int \frac{1}{\sqrt [3]{b \sec (c+d x)}} \, dx}{b}+\int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\\ &=\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}+\frac{(A+4 C) \int (b \sec (c+d x))^{2/3} \, dx}{4 b^2}+\frac{\left (B \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \sqrt [3]{\frac{\cos (c+d x)}{b}} \, dx}{b}\\ &=-\frac{3 B \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{4 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}+\frac{\left ((A+4 C) \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{2/3}} \, dx}{4 b^2}\\ &=-\frac{3 (A+4 C) \cos (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{4 b^2 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{4 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}\\ \end{align*}

Mathematica [C]  time = 1.93147, size = 298, normalized size = 2.04 \[ \frac{\left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac{30 \cos (c+d x) (4 B \cot (c)-A \sin (c+d x))}{d}+\frac{3 i 2^{2/3} e^{-i d x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right )^{2/3} \left (e^{i d x} \left (8 B e^{i (c+d x)} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (c+d x)}\right )-5 \left (-1+e^{2 i c}\right ) (A+4 C) \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{4}{3},-e^{2 i (c+d x)}\right )\right )+40 B e^{i c} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{2}{3},\frac{5}{6},-e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sec ^{\frac{2}{3}}(c+d x)}\right )}{20 (b \sec (c+d x))^{4/3} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((3*I)*2^(2/3)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(2/3)*(1
+ E^((2*I)*(c + d*x)))^(2/3)*(40*B*E^(I*c)*Hypergeometric2F1[-1/6, 2/3, 5/6, -E^((2*I)*(c + d*x))] + E^(I*d*x)
*(-5*(A + 4*C)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/3, 2/3, 4/3, -E^((2*I)*(c + d*x))] + 8*B*E^(I*(c + d*x))
*Hypergeometric2F1[2/3, 5/6, 11/6, -E^((2*I)*(c + d*x))])))/(d*E^(I*d*x)*(-1 + E^((2*I)*c))*Sec[c + d*x]^(2/3)
) - (30*Cos[c + d*x]*(4*B*Cot[c] - A*Sin[c + d*x]))/d))/(20*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*
(b*Sec[c + d*x])^(4/3))

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Maple [F]  time = 0.143, 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{4}{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))^(4/3),x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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