3.67 \(\int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=225 \[ -\frac {3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-1);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)} \]

[Out]

3*C*sec(d*x+c)^(1+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(4+3*m)-3*(C+3*C*m+A*(4+3*m))*hypergeom([1/2, 1/3-1/2*m
],[4/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(-9*m^2-6*m+8)/(sin(d*x+c)^2)^
(1/2)+3*B*hypergeom([1/2, -1/6-1/2*m],[5/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d
/(1+3*m)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 225, 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 = {20, 4047, 3772, 2643, 4046} \[ -\frac {3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-1);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*(4 + 3*m)) - (3*(C + 3*C*m + A*(4 + 3*m))*Hy
pergeometric2F1[1/2, (2 - 3*m)/6, (8 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])^(1/3)*Si
n[c + d*x])/(d*(2 - 3*m)*(4 + 3*m)*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[1/2, (-1 - 3*m)/6, (5 - 3*m)
/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*(1 + 3*m)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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

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Mathematica [C]  time = 7.32, size = 494, normalized size = 2.20 \[ -\frac {3 i 2^{m+\frac {4}{3}} e^{-\frac {1}{3} i d (3 m+1) x} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac {1}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac {1}{3}} \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {e^{i c} \left (\frac {e^{\frac {1}{3} i (3 c+d (3 m+7) x)} \left (2 (3 m+13) (A+2 C) \, _2F_1\left (m+\frac {7}{3},\frac {1}{6} (3 m+7);\frac {1}{6} (3 m+13);-e^{2 i (c+d x)}\right )+A (3 m+7) e^{2 i (c+d x)} \, _2F_1\left (m+\frac {7}{3},\frac {1}{6} (3 m+13);\frac {1}{6} (3 m+19);-e^{2 i (c+d x)}\right )\right )}{(3 m+7) (3 m+13)}+\frac {2 B e^{\frac {1}{3} i d (3 m+4) x} \, _2F_1\left (m+\frac {7}{3},\frac {1}{6} (3 m+4);\frac {m}{2}+\frac {5}{3};-e^{2 i (c+d x)}\right )}{3 m+4}\right )}{d}+\frac {A e^{\frac {1}{3} i x (3 d m+d)} \, _2F_1\left (m+\frac {7}{3},\frac {1}{6} (3 m+1);\frac {1}{6} (3 m+7);-e^{2 i (c+d x)}\right )}{3 d m+d}+\frac {2 B e^{\frac {1}{3} i (9 c+d (3 m+10) x)} \, _2F_1\left (\frac {m}{2}+\frac {5}{3},m+\frac {7}{3};\frac {m}{2}+\frac {8}{3};-e^{2 i (c+d x)}\right )}{d (3 m+10)}\right )}{\sec ^{\frac {7}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((-3*I)*2^(4/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1/3 + m)*(1 + E^((2*I)*(c + d*x)))^(1/3 + m)*
((2*B*E^((I/3)*(9*c + d*(10 + 3*m)*x))*Hypergeometric2F1[5/3 + m/2, 7/3 + m, 8/3 + m/2, -E^((2*I)*(c + d*x))])
/(d*(10 + 3*m)) + (A*E^((I/3)*(d + 3*d*m)*x)*Hypergeometric2F1[7/3 + m, (1 + 3*m)/6, (7 + 3*m)/6, -E^((2*I)*(c
+ d*x))])/(d + 3*d*m) + (E^(I*c)*((2*B*E^((I/3)*d*(4 + 3*m)*x)*Hypergeometric2F1[7/3 + m, (4 + 3*m)/6, 5/3 +
m/2, -E^((2*I)*(c + d*x))])/(4 + 3*m) + (E^((I/3)*(3*c + d*(7 + 3*m)*x))*(2*(A + 2*C)*(13 + 3*m)*Hypergeometri
c2F1[7/3 + m, (7 + 3*m)/6, (13 + 3*m)/6, -E^((2*I)*(c + d*x))] + A*E^((2*I)*(c + d*x))*(7 + 3*m)*Hypergeometri
c2F1[7/3 + m, (13 + 3*m)/6, (19 + 3*m)/6, -E^((2*I)*(c + d*x))]))/((7 + 3*m)*(13 + 3*m))))/d)*(b*Sec[c + d*x])
^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(E^((I/3)*d*(1 + 3*m)*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2
*c + 2*d*x])*Sec[c + d*x]^(7/3))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(1/3)*sec(d*x + c)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(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))^(1/3)*sec(d*x + c)^m, x)

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(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))^(1/3)*sec(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^(1/3)*(1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int((b/cos(c + d*x))^(1/3)*(1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**(1/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((b*sec(c + d*x))**(1/3)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**m, x)

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