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

Optimal. Leaf size=154 \[ \frac {3 (11 A+8 C) \sin (c+d x) (b \sec (c+d x))^{5/3} \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\cos ^2(c+d x)\right )}{55 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{8/3} \, _2F_1\left (-\frac {4}{3},\frac {1}{2};-\frac {1}{3};\cos ^2(c+d x)\right )}{8 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{8/3}}{11 b^2 d} \]

[Out]

3/55*(11*A+8*C)*hypergeom([-5/6, 1/2],[1/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(5/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(
1/2)+3/8*B*hypergeom([-4/3, 1/2],[-1/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(8/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/
2)+3/11*C*(b*sec(d*x+c))^(8/3)*tan(d*x+c)/b^2/d

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Rubi [A]  time = 0.16, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac {3 (11 A+8 C) \sin (c+d x) (b \sec (c+d x))^{5/3} \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\cos ^2(c+d x)\right )}{55 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{8/3} \, _2F_1\left (-\frac {4}{3},\frac {1}{2};-\frac {1}{3};\cos ^2(c+d x)\right )}{8 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{8/3}}{11 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(11*A + 8*C)*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(5/3)*Sin[c + d*x])/(55*b*d
*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[-4/3, 1/2, -1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(8/3)*Sin[c
+ d*x])/(8*b^2*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(11*b^2*d)

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 4046

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]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (b \sec (c+d x))^{8/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac {\int (b \sec (c+d x))^{8/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{11/3} \, dx}{b^3}\\ &=\frac {3 C (b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b^2 d}+\frac {(11 A+8 C) \int (b \sec (c+d x))^{8/3} \, dx}{11 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 )^{11/3}} \, dx}{b^3}\\ &=\frac {3 C (b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b^2 d}+\frac {3 B \, _2F_1\left (-\frac {4}{3},\frac {1}{2};-\frac {1}{3};\cos ^2(c+d x)\right ) \sec (c+d x) (b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d \sqrt {\sin ^2(c+d x)}}+\frac {\left ((11 A+8 C) \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 )^{8/3}} \, dx}{11 b^2}\\ &=\frac {3 (11 A+8 C) \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{55 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b^2 d}+\frac {3 B \, _2F_1\left (-\frac {4}{3},\frac {1}{2};-\frac {1}{3};\cos ^2(c+d x)\right ) \sec (c+d x) (b \sec (c+d x))^{2/3} \tan (c+d x)}{8 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 5.01, size = 346, normalized size = 2.25 \[ \frac {3 (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sec ^{\frac {2}{3}}(c+d x) \left (2 \tan (c+d x) \sec ^2(c+d x) (4 (11 A+8 C) \cos (2 (c+d x))+44 A+55 B \cos (c+d x)+72 C)+275 B \csc (c) \cos (d x)\right )-\frac {i 2^{2/3} e^{-i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (16 \left (-1+e^{2 i c}\right ) (11 A+8 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-e^{2 i (c+d x)}\right )+275 B \left (-1+e^{2 i c}\right ) \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {2}{3};\frac {5}{6};-e^{2 i (c+d x)}\right )+275 B \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{440 d \sec ^{\frac {8}{3}}(c+d x) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((-I)*2^(2/3)*(E^(I*(c + d*x))/(1 + E^((2*I
)*(c + d*x))))^(2/3)*(275*B*(1 + E^((2*I)*(c + d*x))) + 275*B*(-1 + E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^(2/
3)*Hypergeometric2F1[-1/6, 2/3, 5/6, -E^((2*I)*(c + d*x))] + 16*(11*A + 8*C)*E^(I*(c + d*x))*(-1 + E^((2*I)*c)
)*(1 + E^((2*I)*(c + d*x)))^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-
1 + E^((2*I)*c))) + Sec[c + d*x]^(2/3)*(275*B*Cos[d*x]*Csc[c] + 2*(44*A + 72*C + 55*B*Cos[c + d*x] + 4*(11*A +
8*C)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Tan[c + d*x])))/(440*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]
)*Sec[c + d*x]^(8/3))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{4} + B \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.98, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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