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

Optimal. Leaf size=165 \[ \frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-1);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)}}-\frac {3 A \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3*A*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/
(2-3*m)/(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)

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {20, 3787, 3772, 2643} \[ \frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-1);\frac {1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)}}-\frac {3 A \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*Hypergeometric2F1[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)*Sin[c + d*x])/(d*(2 - 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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

\begin {align*} \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\frac {\sqrt [3]{b \sec (c+d x)} \int \sec ^{\frac {1}{3}+m}(c+d x) (A+B \sec (c+d x)) \, dx}{\sqrt [3]{\sec (c+d x)}}\\ &=\frac {\left (A \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {1}{3}+m}(c+d x) \, 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)}}\\ &=\left (A \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+\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 A \, _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) \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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.28, size = 140, normalized size = 0.85 \[ \frac {3 \sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \left (A (3 m+4) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+1);\frac {1}{6} (3 m+7);\sec ^2(c+d x)\right )+B (3 m+1) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+4);\frac {m}{2}+\frac {5}{3};\sec ^2(c+d x)\right )\right )}{d (3 m+1) (3 m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x]*(A*(4 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + 3*m)/6, (7 + 3*m)/6, Sec[c + d*x]^2] + B
*(1 + 3*m)*Hypergeometric2F1[1/2, (4 + 3*m)/6, 5/3 + m/2, Sec[c + d*x]^2])*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/
3)*Sqrt[-Tan[c + d*x]^2])/(d*(1 + 3*m)*(4 + 3*m))

________________________________________________________________________________________

fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.53, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

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)),x)

[Out]

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

________________________________________________________________________________________

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

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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)),x)

[Out]

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

________________________________________________________________________________________