\(\int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [32]

Optimal result

Integrand size = 31, antiderivative size = 173 \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7-3 m),\frac {1}{6} (13-3 m),\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{b d (7-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3*A*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/(7-3*m)/(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*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {20, 3872, 3857, 2722} \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 A \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7-3 m),\frac {1}{6} (13-3 m),\cos ^2(c+d x)\right )}{b d (7-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 B \sin (c+d x) \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[In]

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

[Out]

(-3*A*Hypergeometric2F1[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*(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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 3872

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*} \text {integral}& = \frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac {4}{3}+m}(c+d x) (A+B \sec (c+d x)) \, dx}{b \sqrt [3]{b \sec (c+d x)}} \\ & = \frac {\left (A \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac {4}{3}+m}(c+d x) \, 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 {\left (A \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 \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 A \operatorname {Hypergeometric2F1}\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 (7-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\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*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 \csc (c+d x) \left (A (-1+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4+3 m),\frac {1}{6} (2+3 m),\sec ^2(c+d x)\right )+B (-4+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) \sqrt {-\tan ^2(c+d x)}}{d (-4+3 m) (-1+3 m) (b \sec (c+d x))^{4/3}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int(sec(d*x+c)^m*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)

[Out]

int(sec(d*x+c)^m*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)

Fricas [F]

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

[In]

integrate(sec(d*x+c)^m*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)**m*(A+B*sec(d*x+c))/(b*sec(d*x+c))**(4/3),x)

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)**m/(b*sec(c + d*x))**(4/3), x)

Maxima [F]

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

[In]

integrate(sec(d*x+c)^m*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^m/(b*sec(d*x + c))^(4/3), x)

Giac [F]

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

[In]

integrate(sec(d*x+c)^m*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^m/(b*sec(d*x + c))^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

[In]

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

[Out]

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