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

Optimal. Leaf size=228 \[ \frac{3 (C (1-3 m)-A (3 m+2)) \sin (c+d x) \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (4-3 m),\frac{1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) (3 m+2) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \sec ^m(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (1-3 m),\frac{1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac{3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+2) \sqrt [3]{b \sec (c+d x)}} \]

[Out]

(3*C*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + 3*m)*(b*Sec[c + d*x])^(1/3)) + (3*(C*(1 - 3*m) - A*(2 + 3*m))*
Hypergeometric2F1[1/2, (4 - 3*m)/6, (10 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(4 -
3*m)*(2 + 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/2, (1 - 3*m)/6, (7 - 3*
m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*Sin[c + d*x])/(d*(1 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.190819, antiderivative size = 228, normalized size of antiderivative = 1., 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} \[ \frac{3 (C (1-3 m)-A (3 m+2)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4-3 m);\frac{1}{6} (10-3 m);\cos ^2(c+d x)\right )}{d (4-3 m) (3 m+2) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \sec ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (1-3 m);\frac{1}{6} (7-3 m);\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac{3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+2) \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + 3*m)*(b*Sec[c + d*x])^(1/3)) + (3*(C*(1 - 3*m) - A*(2 + 3*m))*
Hypergeometric2F1[1/2, (4 - 3*m)/6, (10 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(4 -
3*m)*(2 + 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/2, (1 - 3*m)/6, (7 - 3*
m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*Sin[c + d*x])/(d*(1 - 3*m)*(b*Sec[c + d*x])^(1/3)*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 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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx &=\frac{\sqrt [3]{\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]{b \sec (c+d x)}}\\ &=\frac{\sqrt [3]{\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]{b \sec (c+d x)}}+\frac{\left (B \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{\frac{2}{3}+m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}+\frac{\left (\left (C \left (-\frac{1}{3}+m\right )+A \left (\frac{2}{3}+m\right )\right ) \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac{1}{3}+m}(c+d x) \, dx}{\left (\frac{2}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}+\frac{\left (B \cos ^{\frac{2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{-\frac{2}{3}-m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \, _2F_1\left (\frac{1}{2},\frac{1}{6} (1-3 m);\frac{1}{6} (7-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (1-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}+\frac{\left (\left (C \left (-\frac{1}{3}+m\right )+A \left (\frac{2}{3}+m\right )\right ) \cos ^{\frac{2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac{1}{3}-m}(c+d x) \, dx}{\left (\frac{2}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \sec (c+d x)}}+\frac{3 (C (1-3 m)-A (2+3 m)) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4-3 m);\frac{1}{6} (10-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) (2+3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}-\frac{3 B \, _2F_1\left (\frac{1}{2},\frac{1}{6} (1-3 m);\frac{1}{6} (7-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (1-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 11.5551, size = 548, normalized size = 2.4 \[ -\frac{3 i 2^{m+\frac{2}{3}} e^{-\frac{1}{3} i (3 c+d (3 m+2) x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{2}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac{2}{3}} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (e^{i c} (3 m-1) \left ((3 m+2) e^{\frac{1}{3} i (3 c+d (3 m+5) x)} \left ((3 m+5) e^{i (c+d x)} \left (A (3 m+8) e^{i (c+d x)} \text{Hypergeometric2F1}\left (m+\frac{5}{3},\frac{1}{6} (3 m+11),\frac{1}{6} (3 m+17),-e^{2 i (c+d x)}\right )+2 B (3 m+11) \text{Hypergeometric2F1}\left (m+\frac{5}{3},\frac{1}{6} (3 m+8),\frac{m}{2}+\frac{7}{3},-e^{2 i (c+d x)}\right )\right )+2 \left (9 m^2+57 m+88\right ) (A+2 C) \text{Hypergeometric2F1}\left (m+\frac{5}{3},\frac{1}{6} (3 m+5),\frac{1}{6} (3 m+11),-e^{2 i (c+d x)}\right )\right )+2 B \left (27 m^3+216 m^2+549 m+440\right ) e^{\frac{1}{3} i d (3 m+2) x} \text{Hypergeometric2F1}\left (m+\frac{5}{3},\frac{1}{6} (3 m+2),\frac{1}{6} (3 m+8),-e^{2 i (c+d x)}\right )\right )+A \left (81 m^4+702 m^3+2079 m^2+2418 m+880\right ) e^{\frac{1}{3} i d (3 m-1) x} \text{Hypergeometric2F1}\left (m+\frac{5}{3},\frac{1}{6} (3 m-1),\frac{1}{6} (3 m+5),-e^{2 i (c+d x)}\right )\right )}{d (3 m-1) (3 m+2) (3 m+5) (3 m+8) (3 m+11) \sec ^{\frac{5}{3}}(c+d x) \sqrt [3]{b \sec (c+d x)} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-3*I)*2^(2/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(2/3 + m)*(1 + E^((2*I)*(c + d*x)))^(2/3 + m)*
(A*E^((I/3)*d*(-1 + 3*m)*x)*(880 + 2418*m + 2079*m^2 + 702*m^3 + 81*m^4)*Hypergeometric2F1[5/3 + m, (-1 + 3*m)
/6, (5 + 3*m)/6, -E^((2*I)*(c + d*x))] + E^(I*c)*(-1 + 3*m)*(2*B*E^((I/3)*d*(2 + 3*m)*x)*(440 + 549*m + 216*m^
2 + 27*m^3)*Hypergeometric2F1[5/3 + m, (2 + 3*m)/6, (8 + 3*m)/6, -E^((2*I)*(c + d*x))] + E^((I/3)*(3*c + d*(5
+ 3*m)*x))*(2 + 3*m)*(2*(A + 2*C)*(88 + 57*m + 9*m^2)*Hypergeometric2F1[5/3 + m, (5 + 3*m)/6, (11 + 3*m)/6, -E
^((2*I)*(c + d*x))] + E^(I*(c + d*x))*(5 + 3*m)*(2*B*(11 + 3*m)*Hypergeometric2F1[5/3 + m, (8 + 3*m)/6, 7/3 +
m/2, -E^((2*I)*(c + d*x))] + A*E^(I*(c + d*x))*(8 + 3*m)*Hypergeometric2F1[5/3 + m, (11 + 3*m)/6, (17 + 3*m)/6
, -E^((2*I)*(c + d*x))]))))*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^((I/3)*(3*c + d*(2 + 3*m)*x))*(-1 +
3*m)*(2 + 3*m)*(5 + 3*m)*(8 + 3*m)*(11 + 3*m)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(
5/3)*(b*Sec[c + d*x])^(1/3))

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Maple [F]  time = 0.179, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt [3]{b\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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 )^{m}}{b \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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 )}}{\sqrt [3]{b \sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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