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

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

[Out]

(3*b*(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*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*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(4/3)*Tan[c + d*x])/(7*d)

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Rubi [A]  time = 0.137347, 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, 4046} \[ \frac{3 b (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(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*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*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(4/3)*Tan[c + d*x])/(7*d)

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 (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{B \int (b \sec (c+d x))^{7/3} \, dx}{b}+\int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 d}+\frac{1}{7} (7 A+4 C) \int (b \sec (c+d x))^{4/3} \, dx+\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}\\ &=\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 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 d}+\frac{1}{7} \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\\ &=\frac{3 b (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 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 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 d}\\ \end{align*}

Mathematica [C]  time = 2.39704, size = 290, normalized size = 1.99 \[ -\frac{3 i b \sqrt [3]{b \sec (c+d x)} \left (-14 A e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{7/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{5}{3},-e^{2 i (c+d x)}\right )+7 B \left (1+e^{2 i (c+d x)}\right )^{7/3} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (c+d x)}\right )-8 C e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{7/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{5}{3},-e^{2 i (c+d x)}\right )+28 A e^{i (c+d x)}+56 A e^{3 i (c+d x)}+28 A e^{5 i (c+d x)}+7 B e^{4 i (c+d x)}-7 B+8 C e^{i (c+d x)}+40 C e^{3 i (c+d x)}+16 C e^{5 i (c+d x)}\right )}{28 d \left (1+e^{2 i (c+d x)}\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/28)*b*(-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))] - 14*A*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))] - 8*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))])*(b*Sec[c + d*x])^(1/3))/(d*(1 + E^((2
*I)*(c + d*x)))^2)

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Maple [F]  time = 0.139, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \sec \left (d x + c\right )^{3} + B b \sec \left (d x + c\right )^{2} + A b \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{1}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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), x)