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

Optimal. Leaf size=144 \[ \frac {3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)}-\frac {3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt {\sin ^2(c+d x)}} \]

[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)

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Rubi [A]  time = 0.12, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {20, 4046, 3772, 2643} \[ \frac {3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)}-\frac {3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

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]

Rubi steps

\begin {align*} \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+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+C \sec ^2(c+d x)\right ) \, 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)}}\\ &=\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 ) \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)}}\\ \end {align*}

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Mathematica [C]  time = 2.85, size = 303, normalized size = 2.10 \[ -\frac {3 i 2^{m+\frac {4}{3}} e^{-\frac {1}{3} i (3 m+4) (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac {4}{3}} \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {2 (A+2 C) e^{\frac {1}{3} i (3 m+7) (c+d x)} \, _2F_1\left (1,\frac {1}{6} (-3 m-1);\frac {1}{6} (3 m+13);-e^{2 i (c+d x)}\right )}{3 m+7}+\frac {A e^{\frac {1}{3} i (3 m+1) (c+d x)} \, _2F_1\left (1,\frac {1}{6} (-3 m-7);\frac {1}{6} (3 m+7);-e^{2 i (c+d x)}\right )}{3 m+1}+\frac {A e^{\frac {1}{3} i (3 m+13) (c+d x)} \, _2F_1\left (1,\frac {1}{6} (5-3 m);\frac {1}{6} (3 m+19);-e^{2 i (c+d x)}\right )}{3 m+13}\right )}{d \sec ^{\frac {7}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A + 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))))^(4/3 + m)*((A*E^((I/3)*(1 + 3*m)*(c + d*x))*Hy
pergeometric2F1[1, (-7 - 3*m)/6, (7 + 3*m)/6, -E^((2*I)*(c + d*x))])/(1 + 3*m) + (2*(A + 2*C)*E^((I/3)*(7 + 3*
m)*(c + d*x))*Hypergeometric2F1[1, (-1 - 3*m)/6, (13 + 3*m)/6, -E^((2*I)*(c + d*x))])/(7 + 3*m) + (A*E^((I/3)*
(13 + 3*m)*(c + d*x))*Hypergeometric2F1[1, (5 - 3*m)/6, (19 + 3*m)/6, -E^((2*I)*(c + d*x))])/(13 + 3*m))*(b*Se
c[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2))/(d*E^((I/3)*(4 + 3*m)*(c + d*x))*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[
c + d*x]^(7/3))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m}, x\right ) \]

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

[Out]

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

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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.44, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

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+C*sec(d*x+c)^2),x)

[Out]

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

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

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

[Out]

integrate((C*sec(d*x + c)^2 + 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 = -0.01 \[ \int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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+C*sec(d*x+c)**2),x)

[Out]

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

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