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

Optimal. Leaf size=119 \[ \frac {3 A \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \, _2F_1\left (-\frac {5}{3},\frac {1}{2};-\frac {2}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/7*A*hypergeom([-7/6, 1/2],[-1/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(7/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/10
*B*hypergeom([-5/3, 1/2],[-2/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(10/3)*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 = 119, 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} \[ \frac {3 A \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \, _2F_1\left (-\frac {5}{3},\frac {1}{2};-\frac {2}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c + d*x])/(7*b*d*Sqrt[Sin[c
+ d*x]^2]) + (3*B*Hypergeometric2F1[-5/3, 1/2, -2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(10/3)*Sin[c + d*x])/(1
0*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 \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx &=\frac {\int (b \sec (c+d x))^{10/3} (A+B \sec (c+d x)) \, dx}{b^2}\\ &=\frac {A \int (b \sec (c+d x))^{10/3} \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{13/3} \, dx}{b^3}\\ &=\frac {\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 )^{10/3}} \, dx}{b^2}+\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 )^{13/3}} \, dx}{b^3}\\ &=\frac {3 A b \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \, _2F_1\left (-\frac {5}{3},\frac {1}{2};-\frac {2}{3};\cos ^2(c+d x)\right ) \sec ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{10 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 90, normalized size = 0.76 \[ -\frac {3 \left (-\tan ^2(c+d x)\right )^{3/2} \csc ^3(c+d x) (b \sec (c+d x))^{4/3} \left (13 A \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\sec ^2(c+d x)\right )+10 B \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\sec ^2(c+d x)\right )\right )}{130 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Csc[c + d*x]^3*(13*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3, 8/3, Sec[c + d*x]^2] + 10*B*Hypergeometric2F
1[1/2, 13/6, 19/6, Sec[c + d*x]^2])*(b*Sec[c + d*x])^(4/3)*(-Tan[c + d*x]^2)^(3/2))/(130*d)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b \sec \left (d x + c\right )^{4} + A b \sec \left (d x + c\right )^{3}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b*sec(d*x + c)^4 + A*b*sec(d*x + c)^3)*(b*sec(d*x + c))^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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)*sec(d*x + c)^2, x)

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maple [F]  time = 0.77, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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)*sec(d*x + c)^2, x)

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

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))/cos(c + d*x)^2,x)

[Out]

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

[Out]

Timed out

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