Integrand size = 31, antiderivative size = 119 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/3} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{10/3} \sin (c+d x)}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \] Output:
3/7*A*hypergeom([-7/6, 1/2],[-1/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(7/3)*sin( d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/10*B*hypergeom([-5/3, 1/2],[-2/3],cos(d* x+c)^2)*(b*sec(d*x+c))^(10/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=-\frac {3 \csc ^3(c+d x) \left (13 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right )+10 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{4/3} \left (-\tan ^2(c+d x)\right )^{3/2}}{130 d} \] Input:
Integrate[Sec[c + d*x]^2*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x]),x]
Output:
(-3*Csc[c + d*x]^3*(13*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3, 8/3, Sec [c + d*x]^2] + 10*B*Hypergeometric2F1[1/2, 13/6, 19/6, Sec[c + d*x]^2])*(b *Sec[c + d*x])^(4/3)*(-Tan[c + d*x]^2)^(3/2))/(130*d)
Time = 0.53 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2030, 3042, 4274, 3042, 4259, 3042, 3122}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{10/3} (A+B \sec (c+d x))dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b^2}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {A \int (b \sec (c+d x))^{10/3}dx+\frac {B \int (b \sec (c+d x))^{13/3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{13/3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{10/3}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{13/3}}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{10/3}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{13/3}}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {3 A b \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}}{b^2}\) |
Input:
Int[Sec[c + d*x]^2*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x]),x]
Output:
((3*A*b*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x] )^(7/3)*Sin[c + d*x])/(7*d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[ -5/3, 1/2, -2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(10/3)*Sin[c + d*x])/(10 *d*Sqrt[Sin[c + d*x]^2]))/b^2
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
\[\int \sec \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]
Input:
int(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x)
Output:
int(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x)
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x, algorithm= "fricas")
Output:
integral((B*b*sec(d*x + c)^4 + A*b*sec(d*x + c)^3)*(b*sec(d*x + c))^(1/3), x)
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**2*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)),x)
Output:
Timed out
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*sec(d*x + c)^2, x)
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*sec(d*x + c)^2, x)
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:
int(((A + B/cos(c + d*x))*(b/cos(c + d*x))^(4/3))/cos(c + d*x)^2,x)
Output:
int(((A + B/cos(c + d*x))*(b/cos(c + d*x))^(4/3))/cos(c + d*x)^2, x)
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=b^{\frac {4}{3}} \left (\left (\int \sec \left (d x +c \right )^{\frac {13}{3}}d x \right ) b +\left (\int \sec \left (d x +c \right )^{\frac {10}{3}}d x \right ) a \right ) \] Input:
int(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x)
Output:
b**(1/3)*b*(int(sec(c + d*x)**(1/3)*sec(c + d*x)**4,x)*b + int(sec(c + d*x )**(1/3)*sec(c + d*x)**3,x)*a)