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

Optimal. Leaf size=91 \[ \frac {3 b C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d}-\frac {3 b^2 (4 A+C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

[Out]

-3/8*b^2*(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)+3/4*b*C*(b*sec(d*x+c))^(1/3)*tan(d*x+c)/d

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Rubi [A]  time = 0.09, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 4046, 3772, 2643} \[ \frac {3 b C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d}-\frac {3 b^2 (4 A+C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*(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)*S
qrt[Sin[c + d*x]^2]) + (3*b*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*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]

Rubi steps

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

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Mathematica [C]  time = 1.50, size = 163, normalized size = 1.79 \[ \frac {3 b \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (C \sin (c+d x) \sec ^{\frac {4}{3}}(c+d x)-i \sqrt [3]{2} (4 A+C) \sqrt [3]{\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (c+d x)}\right )\right )}{2 d \sec ^{\frac {7}{3}}(c+d x) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2)*((-I)*2^(1/3)*(4*A + C)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c +
d*x))))^(1/3)*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(c + d*x))] + C*Sec[
c + d*x]^(4/3)*Sin[c + d*x]))/(2*d*(A + 2*C + A*Cos[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 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} + A b \cos \left (d x + c\right ) \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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)*cos(d*x + c), x)

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maple [F]  time = 2.04, size = 0, normalized size = 0.00 \[ \int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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)*cos(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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