3.1.17 \(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [17]

Optimal. Leaf size=92 \[ -\frac {3 (2 A-C) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}} \]

[Out]

-3/8*(2*A-C)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)+3
/2*C*tan(d*x+c)/b/d/(b*sec(d*x+c))^(1/3)

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Rubi [A]
time = 0.05, antiderivative size = 92, 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 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}}-\frac {3 (2 A-C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(2*A - C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(4/3)*Sqrt[
Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(2*b*d*(b*Sec[c + d*x])^(1/3))

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 \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx &=\frac {\int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx}{b}\\ &=\frac {3 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}}+\frac {(2 A-C) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}} \, dx}{2 b}\\ &=\frac {3 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}}+\frac {\left ((2 A-C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \sqrt [3]{\frac {\cos (c+d x)}{b}} \, dx}{2 b}\\ &=-\frac {3 (2 A-C) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.13, size = 130, normalized size = 1.41 \begin {gather*} -\frac {3 i \left (-5 \left (A+A e^{2 i (c+d x)}-C e^{2 i (c+d x)}\right )+(2 A-C) e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (c+d x)}\right )\right )}{5 b d \left (1+e^{2 i (c+d x)}\right ) \sqrt [3]{b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/5)*(-5*(A + A*E^((2*I)*(c + d*x)) - C*E^((2*I)*(c + d*x))) + (2*A - C)*E^((2*I)*(c + d*x))*(1 + E^((2
*I)*(c + d*x)))^(2/3)*Hypergeometric2F1[2/3, 5/6, 11/6, -E^((2*I)*(c + d*x))]))/(b*d*(1 + E^((2*I)*(c + d*x)))
*(b*Sec[c + d*x])^(1/3))

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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \frac {\sec \left (d x +c \right ) \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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