3.14 \(\int (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=112 \[ \frac {3 A b \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*A*b*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)+3/4*B*h
ypergeom([-2/3, 1/2],[1/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(4/3)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3787, 3772, 2643} \[ \frac {3 A b \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*Sqrt[Sin[c +
d*x]^2]) + (3*B*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(4/3)*Sin[c + d*x])/(4*d*Sq
rt[Sin[c + d*x]^2])

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

\begin {align*} \int (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx &=A \int (b \sec (c+d x))^{4/3} \, dx+\frac {B \int (b \sec (c+d x))^{7/3} \, dx}{b}\\ &=\left (A \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 {\left (B \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 )^{7/3}} \, dx}{b}\\ &=\frac {3 A b \, _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)}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 88, normalized size = 0.79 \[ \frac {3 \sqrt {-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^{4/3} \left (7 A \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\sec ^2(c+d x)\right )+4 B \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\sec ^2(c+d x)\right )\right )}{28 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x]*(7*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Sec[c + d*x]^2] + 4*B*Hypergeometric2F1[1/2
, 7/6, 13/6, Sec[c + d*x]^2])*(b*Sec[c + d*x])^(4/3)*Sqrt[-Tan[c + d*x]^2])/(28*d)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b \sec \left (d x + c\right )^{2} + A b \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b*sec(d*x + c)^2 + A*b*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 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.98, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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