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

Optimal. Leaf size=234 \[ -\frac {3 (-3 A m+A+C (4-3 m)) \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (7-3 m);\frac {1}{6} (13-3 m);\cos ^2(c+d x)\right )}{b d (1-3 m) (7-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 B \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 C \sin (c+d x) \sec ^m(c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}} \]

[Out]

-3*C*sec(d*x+c)^m*sin(d*x+c)/b/d/(1-3*m)/(b*sec(d*x+c))^(1/3)-3*(A+C*(4-3*m)-3*A*m)*hypergeom([1/2, 7/6-1/2*m]
,[13/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-2+m)*sin(d*x+c)/b/d/(9*m^2-24*m+7)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2
)^(1/2)-3*B*hypergeom([1/2, 2/3-1/2*m],[5/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d*x+c)/b/d/(4-3*m)/(b*s
ec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 234, 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} \[ -\frac {3 (-3 A m+A+C (4-3 m)) \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (7-3 m);\frac {1}{6} (13-3 m);\cos ^2(c+d x)\right )}{b d (1-3 m) (7-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 B \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 C \sin (c+d x) \sec ^m(c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}} \]

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])^(4/3),x]

[Out]

(-3*C*Sec[c + d*x]^m*Sin[c + d*x])/(b*d*(1 - 3*m)*(b*Sec[c + d*x])^(1/3)) - (3*(A + C*(4 - 3*m) - 3*A*m)*Hyper
geometric2F1[1/2, (7 - 3*m)/6, (13 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-2 + m)*Sin[c + d*x])/(b*d*(1 - 3*m
)*(7 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*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])/(b*d*(4 - 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 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 \frac {\sec ^m(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 {\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac {4}{3}+m}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=\frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac {4}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \sec (c+d x)}}+\frac {\left (B \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac {1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac {3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}+\frac {\left (\left (C \left (-\frac {4}{3}+m\right )+A \left (-\frac {1}{3}+m\right )\right ) \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac {4}{3}+m}(c+d x) \, dx}{b \left (-\frac {1}{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 {1}{3}-m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac {3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}-\frac {3 B \, _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)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (C \left (-\frac {4}{3}+m\right )+A \left (-\frac {1}{3}+m\right )\right ) \cos ^{\frac {2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac {4}{3}-m}(c+d x) \, dx}{b \left (-\frac {1}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac {3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}-\frac {3 (A (1-3 m)+C (4-3 m)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (7-3 m);\frac {1}{6} (13-3 m);\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{b d (1-3 m) (7-3 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} (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)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m}}{b^{2} \sec \left (d x + c\right )^{2}}, x\right ) \]

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))^(4/3),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^2*sec(d*x + c)^2), x
)

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

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))^(4/3),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))^(4/3), x)

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maple [F]  time = 1.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{m}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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))^(4/3),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))^(4/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(b/cos(c + d*x))^(4/3),x)

[Out]

int(((1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(b/cos(c + d*x))^(4/3), x)

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

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))**(4/3),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))**(4/3), x)

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