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

Optimal. Leaf size=93 \[ \frac {3 A b^2 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}-\frac {3 b (2 A+5 C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

[Out]

-3/10*b*(2*A+5*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)+3/5*A*b^2*tan(d*x+c)/d/(b*sec(d*x+c))^(5/3)

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Rubi [A]  time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {16, 4045, 3772, 2643} \[ \frac {3 A b^2 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}-\frac {3 b (2 A+5 C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(2*A + 5*C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*d*(b*Sec[c + d*x])^(2/3)*
Sqrt[Sin[c + d*x]^2]) + (3*A*b^2*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3))

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 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 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 \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=b^2 \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/3}} \, dx\\ &=\frac {3 A b^2 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}+\frac {1}{5} (2 A+5 C) \int \sqrt [3]{b \sec (c+d x)} \, dx\\ &=\frac {3 A b^2 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}+\frac {1}{5} \left ((2 A+5 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 (2 A+5 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)}{10 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^2 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 89, normalized size = 0.96 \[ -\frac {3 \sqrt {-\tan ^2(c+d x)} \cot (c+d x) \sqrt [3]{b \sec (c+d x)} \left (A \cos ^2(c+d x) \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\sec ^2(c+d x)\right )-5 C \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sec ^2(c+d x)\right )\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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)^2, x)

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maple [F]  time = 3.27, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)^2*(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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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