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

Optimal. Leaf size=90 \[ \frac {3 b (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d} \]

[Out]

3/7*b*(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)/d/(sin(d*x+c)^2)^(1/
2)+3/7*C*(b*sec(d*x+c))^(4/3)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4046, 3772, 2643} \[ \frac {3 b (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(4/3)*(A + 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*C*(b*Sec[c + d*x])^(4/3)*Tan[c + d*x])/(7*d)

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]

Rubi steps

\begin {align*} \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 {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 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.32, size = 182, normalized size = 2.02 \[ \frac {3 i e^{i (c+d x)} \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left ((7 A+4 C) \left (1+e^{2 i (c+d x)}\right )^{7/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-e^{2 i (c+d x)}\right )-14 A \left (1+e^{2 i (c+d x)}\right )^2-4 C \left (5 e^{2 i (c+d x)}+2 e^{4 i (c+d x)}+1\right )\right ) \left (A+C \sec ^2(c+d x)\right )}{7 d \left (1+e^{2 i (c+d x)}\right )^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/7)*E^(I*(c + d*x))*Cos[c + d*x]^3*(-14*A*(1 + E^((2*I)*(c + d*x)))^2 - 4*C*(1 + 5*E^((2*I)*(c + d*x))
+ 2*E^((4*I)*(c + d*x))) + (7*A + 4*C)*(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])^(4/3)*(A + C*Sec[c + d*x]^2))/(d*(1 + E^((2*I)*(c + d*x)))^2*(A + 2*C + A*Co
s[2*(c + d*x)]))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.84, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________