3.22 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=117 \[ \frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \sin (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[Out]

-3*A*hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)+3/2*B*h
ypergeom([-1/3, 1/2],[2/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 3787, 3772, 2643} \[ \frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \sin (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx &=\frac {\int (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx}{b^2}\\ &=\frac {A \int (b \sec (c+d x))^{2/3} \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{5/3} \, dx}{b^3}\\ &=\frac {\left (A \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{2/3}} \, dx}{b^2}+\frac {\left (B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{5/3}} \, dx}{b^3}\\ &=\frac {3 B \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{2 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 91, normalized size = 0.78 \[ \frac {3 \sqrt {-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^{2/3} \left (5 A \cos (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\sec ^2(c+d x)\right )+2 B \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\sec ^2(c+d x)\right )\right )}{10 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x]*(5*A*Cos[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Sec[c + d*x]^2] + 2*B*Hypergeometric2F1[1/2
, 5/6, 11/6, Sec[c + d*x]^2])*(b*Sec[c + d*x])^(2/3)*Sqrt[-Tan[c + d*x]^2])/(10*b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{2}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)**2/(b*sec(c + d*x))**(4/3), x)

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