3.1.64 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [64]

Optimal. Leaf size=154 \[ \frac {3 (8 A+5 C) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{16 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{5 b^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 d} \]

[Out]

3/16*(8*A+5*C)*hypergeom([-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)+3/5*B*hypergeom([-5/6, 1/2],[1/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(5/3)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/
2)+3/8*C*(b*sec(d*x+c))^(5/3)*tan(d*x+c)/b^3/d

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Rubi [A]
time = 0.11, 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, 4132, 3857, 2722, 4131} \begin {gather*} \frac {3 (8 A+5 C) \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 )}{16 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \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 )}{5 b^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(8*A + 5*C)*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(16*b^2*
d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(5/3)*Sin[c
+ d*x])/(5*b^3*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*b^3*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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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 4132

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 \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx &=\frac {\int (b \sec (c+d x))^{5/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^3}\\ &=\frac {\int (b \sec (c+d x))^{5/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^3}+\frac {B \int (b \sec (c+d x))^{8/3} \, dx}{b^4}\\ &=\frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 d}+\frac {(8 A+5 C) \int (b \sec (c+d x))^{5/3} \, dx}{8 b^3}+\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 )^{8/3}} \, dx}{b^4}\\ &=\frac {3 B \, _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)}{5 b^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 d}+\frac {\left ((8 A+5 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 )^{5/3}} \, dx}{8 b^3}\\ &=\frac {3 (8 A+5 C) \, _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)}{16 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \, _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)}{5 b^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 5.84, size = 348, normalized size = 2.26 \begin {gather*} \frac {3 e^{-i c} \left (-1+e^{2 i c}\right ) \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \csc (c) \left (16 B-40 A e^{i (c+d x)}-5 C e^{i (c+d x)}-80 A e^{3 i (c+d x)}-70 C e^{3 i (c+d x)}-16 B e^{4 i (c+d x)}-40 A e^{5 i (c+d x)}-25 C e^{5 i (c+d x)}-16 B \left (1+e^{2 i (c+d x)}\right )^{8/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-e^{2 i (c+d x)}\right )+(8 A+5 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{8/3} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (c+d x)}\right )\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{40 \sqrt [3]{2} d \left (1+e^{2 i (c+d x)}\right )^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {2}{3}}(c+d x) (b \sec (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-1 + E^((2*I)*c))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(2/3)*Csc[c]*(16*B - 40*A*E^(I*(c + d*x)) -
5*C*E^(I*(c + d*x)) - 80*A*E^((3*I)*(c + d*x)) - 70*C*E^((3*I)*(c + d*x)) - 16*B*E^((4*I)*(c + d*x)) - 40*A*E^
((5*I)*(c + d*x)) - 25*C*E^((5*I)*(c + d*x)) - 16*B*(1 + E^((2*I)*(c + d*x)))^(8/3)*Hypergeometric2F1[1/3, 2/3
, 4/3, -E^((2*I)*(c + d*x))] + (8*A + 5*C)*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(8/3)*Hypergeometric2F1[2
/3, 5/6, 11/6, -E^((2*I)*(c + d*x))])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(40*2^(1/3)*d*E^(I*c)*(1 + E^((
2*I)*(c + d*x)))^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(2/3)*(b*Sec[c + d*x])^(4/3)
)

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Maple [F]
time = 0.31, size = 0, normalized size = 0.00 \[\int \frac {\left (\sec ^{3}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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