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

Optimal. Leaf size=154 \[ \frac {3 (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )}{91 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \, _2F_1\left (-\frac {5}{3},\frac {1}{2};-\frac {2}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 b^2 d} \]

[Out]

3/91*(13*A+10*C)*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)+3/13*C*(b*sec(d*x+c))^(10/3)*tan(d*x+c)/b^2/d

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Rubi [A]  time = 0.15, 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} \[ \frac {3 (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )}{91 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \, _2F_1\left (-\frac {5}{3},\frac {1}{2};-\frac {2}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(13*A + 10*C)*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c + d*x])/(91*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])/(10*b^2*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(10/3)*Tan[c + d*x])/(13*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))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (b \sec (c+d x))^{10/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac {\int (b \sec (c+d x))^{10/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{13/3} \, dx}{b^3}\\ &=\frac {3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac {(13 A+10 C) \int (b \sec (c+d x))^{10/3} \, dx}{13 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 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\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)}}+\frac {\left ((13 A+10 C) \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}{13 b^2}\\ &=\frac {3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac {3 b (13 A+10 C) \, _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)}{91 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 [C]  time = 6.63, size = 444, normalized size = 2.88 \[ \frac {3 b \csc (c) e^{-i d x} \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (40 \sqrt [3]{2} \left (-1+e^{2 i c}\right ) (13 A+10 C) e^{2 i d x} \sqrt [3]{\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-e^{2 i (c+d x)}\right )-\frac {\left (-1+e^{2 i c}\right ) e^{-i (c-d x)} \sqrt [3]{\sec (c+d x)} \left (80 e^{i (c+d x)} \left (13 A \left (5 e^{2 i (c+d x)}+2 e^{4 i (c+d x)}+1\right ) \left (1+e^{2 i (c+d x)}\right )^2+2 C \left (21 e^{2 i (c+d x)}+79 e^{4 i (c+d x)}+45 e^{6 i (c+d x)}+10 e^{8 i (c+d x)}+5\right )\right )+637 B \left (1+e^{2 i (c+d x)}\right )^{13/3} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (c+d x)}\right )+91 B \left (-30 e^{2 i (c+d x)}+30 e^{6 i (c+d x)}+7 e^{8 i (c+d x)}-7\right )\right )}{2 \left (1+e^{2 i (c+d x)}\right )^4}\right )}{1820 d \sec ^{\frac {7}{3}}(c+d x) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*Csc[c]*(40*2^(1/3)*(13*A + 10*C)*E^((2*I)*d*x)*(-1 + E^((2*I)*c))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x
))))^(1/3)*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -E^((2*I)*(c + d*x))] - ((-1 + E^(
(2*I)*c))*(91*B*(-7 - 30*E^((2*I)*(c + d*x)) + 30*E^((6*I)*(c + d*x)) + 7*E^((8*I)*(c + d*x))) + 80*E^(I*(c +
d*x))*(13*A*(1 + E^((2*I)*(c + d*x)))^2*(1 + 5*E^((2*I)*(c + d*x)) + 2*E^((4*I)*(c + d*x))) + 2*C*(5 + 21*E^((
2*I)*(c + d*x)) + 79*E^((4*I)*(c + d*x)) + 45*E^((6*I)*(c + d*x)) + 10*E^((8*I)*(c + d*x)))) + 637*B*(1 + E^((
2*I)*(c + d*x)))^(13/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^(1/3))/(2*E^(I*(c
- d*x))*(1 + E^((2*I)*(c + d*x)))^4))*(b*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(1820*d
*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(7/3))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \sec \left (d x + c\right )^{5} + B b \sec \left (d x + c\right )^{4} + A b \sec \left (d x + c\right )^{3}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b*sec(d*x + c)^5 + 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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{2}\,{d x} \]

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)+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))^(4/3)*sec(d*x + c)^2, x)

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maple [F]  time = 0.92, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [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)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F(-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)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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