3.1.4 \(\int \cos (c+d x) \sqrt [3]{b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [4]

Optimal. Leaf size=89 \[ -\frac {3 b^2 (A-2 C) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \sec (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \]

[Out]

-3/5*b^2*(A-2*C)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(5/3)/(sin(d*x+c)^2)^(1
/2)+3*b*C*tan(d*x+c)/d/(b*sec(d*x+c))^(2/3)

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Rubi [A]
time = 0.06, antiderivative size = 89, 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, 4131, 3857, 2722} \begin {gather*} \frac {3 b C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}-\frac {3 b^2 (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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

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

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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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

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Mathematica [A]
time = 0.10, size = 93, normalized size = 1.04 \begin {gather*} -\frac {3 \cot (c+d x) \left (2 A \cos ^2(c+d x) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\sec ^2(c+d x)\right )-C \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{4/3} \sqrt {-\tan ^2(c+d x)}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

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

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Maple [F]
time = 0.45, size = 0, normalized size = 0.00 \[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x)

[Out]

int(cos(d*x+c)*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)*sec(d*x + c)^2 + A*cos(d*x + c))*(b*sec(d*x + c))^(1/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{b \sec {\left (c + d x \right )}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(c + d*x))**(1/3)*(A + C*sec(c + d*x)**2)*cos(c + d*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )}^{1/3} \,d x \end {gather*}

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))^(1/3),x)

[Out]

int(cos(c + d*x)*(A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(1/3), x)

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