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

Optimal. Leaf size=165 \[ -\frac {3 A \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 B \sin (c+d x) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[Out]

-3*A*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)/d/(4-3*m)/(b*sec(d*x+c)
)^(1/3)/(sin(d*x+c)^2)^(1/2)-3*B*hypergeom([1/2, 1/6-1/2*m],[7/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*sin(d*x+c)/
d/(1-3*m)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.25, size = 140, normalized size = 0.85 \[ \frac {3 \sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec ^m(c+d x) \left (A (3 m+2) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m-1);\frac {1}{6} (3 m+5);\sec ^2(c+d x)\right )+B (3 m-1) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+2);\frac {1}{6} (3 m+8);\sec ^2(c+d x)\right )\right )}{d (3 m-1) (3 m+2) \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x]*(A*(2 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Sec[c + d*x]^2] +
B*(-1 + 3*m)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Sec[c + d*x]^2])*Sec[c + d*x]^m*Sqrt[-Tan[c + d*
x]^2])/(d*(-1 + 3*m)*(2 + 3*m)*(b*Sec[c + d*x])^(1/3))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m}}{b \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{m}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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