\(\int (f x)^m (d+e x^2) (a+b \sec ^{-1}(c x)) \, dx\) [163]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b e x^{2+m} \sqrt {-1+c^2 x^2}}{c \left (6+5 m+m^2\right ) \sqrt {c^2 x^2}}+\frac {d x^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{1+m}+\frac {e x^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{3+m}+\frac {b \left (e (1+m)^2+c^2 d (2+m) (3+m)\right ) x^{2+m} \sqrt {-1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c (1+m)^2 (2+m) (3+m) \sqrt {c^2 x^2}} \]

[Out]

d*x^(1+m)*(a+b*arcsec(c*x))/(1+m)+e*x^(3+m)*(a+b*arcsec(c*x))/(3+m)-b*e*x^(2+m)*(c^2*x^2-1)^(1/2)/c/(m^2+5*m+6
)/(c^2*x^2)^(1/2)+b*(e*(1+m)^2+c^2*d*(2+m)*(3+m))*x^(2+m)*hypergeom([1, 1+1/2*m],[3/2+1/2*m],c^2*x^2)*(c^2*x^2
-1)^(1/2)/c/(1+m)^2/(2+m)/(3+m)/(c^2*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {14, 5346, 12, 470, 372, 371} \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {d (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}-\frac {b c x \sqrt {1-c^2 x^2} (f x)^{m+1} \left (\frac {e}{c^2 (m+2) (m+3)}+\frac {d}{(m+1)^2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right )}{f \sqrt {c^2 x^2} \sqrt {c^2 x^2-1}}-\frac {b e x \sqrt {c^2 x^2-1} (f x)^{m+1}}{c f \left (m^2+5 m+6\right ) \sqrt {c^2 x^2}} \]

[In]

Int[(f*x)^m*(d + e*x^2)*(a + b*ArcSec[c*x]),x]

[Out]

-((b*e*x*(f*x)^(1 + m)*Sqrt[-1 + c^2*x^2])/(c*f*(6 + 5*m + m^2)*Sqrt[c^2*x^2])) + (d*(f*x)^(1 + m)*(a + b*ArcS
ec[c*x]))/(f*(1 + m)) + (e*(f*x)^(3 + m)*(a + b*ArcSec[c*x]))/(f^3*(3 + m)) - (b*c*(d/(1 + m)^2 + e/(c^2*(2 +
m)*(3 + m)))*x*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(f*Sqrt[
c^2*x^2]*Sqrt[-1 + c^2*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 5346

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[b*c*(x/Sqrt[c^2*x^2]), Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {d (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{(1+m) (3+m) \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}} \\ & = \frac {d (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{\sqrt {-1+c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt {c^2 x^2}} \\ & = -\frac {b e x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {\left (b c \left (-\frac {e (1+m)^2}{c^2 (2+m)}-d (3+m)\right ) x\right ) \int \frac {(f x)^m}{\sqrt {-1+c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt {c^2 x^2}} \\ & = -\frac {b e x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {\left (b c \left (-\frac {e (1+m)^2}{c^2 (2+m)}-d (3+m)\right ) x \sqrt {1-c^2 x^2}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \\ & = -\frac {b e x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {b c \left (\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) x (f x)^{1+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{f (1+m) \left (3+4 m+m^2\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=x (f x)^m \left (\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(1+m)^2 \sqrt {1-c^2 x^2}}+\frac {\frac {(3+m) \left (d (3+m)+e (1+m) x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{1+m}+\frac {b c e \sqrt {1-\frac {1}{c^2 x^2}} x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{\sqrt {1-c^2 x^2}}}{(3+m)^2}\right ) \]

[In]

Integrate[(f*x)^m*(d + e*x^2)*(a + b*ArcSec[c*x]),x]

[Out]

x*(f*x)^m*((b*c*d*Sqrt[1 - 1/(c^2*x^2)]*x*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*Sq
rt[1 - c^2*x^2]) + (((3 + m)*(d*(3 + m) + e*(1 + m)*x^2)*(a + b*ArcSec[c*x]))/(1 + m) + (b*c*e*Sqrt[1 - 1/(c^2
*x^2)]*x^3*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, c^2*x^2])/Sqrt[1 - c^2*x^2])/(3 + m)^2)

Maple [F]

\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )d x\]

[In]

int((f*x)^m*(e*x^2+d)*(a+b*arcsec(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)*(a+b*arcsec(c*x)),x)

Fricas [F]

\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsec(c*x))*(f*x)^m, x)

Sympy [F]

\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

[In]

integrate((f*x)**m*(e*x**2+d)*(a+b*asec(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*asec(c*x))*(d + e*x**2), x)

Maxima [F]

\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

a*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d/(f*(m + 1)) + (((b*e*f^m*m + b*e*f^m)*x^3 + (b*d*f^m*m + 3*b*d*f^m
)*x)*x^m*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - (m^2 + 4*m + 3)*integrate((b*d*f^m*m + 3*b*d*f^m + (b*e*f^m*m +
b*e*f^m)*x^2)*sqrt(c*x + 1)*sqrt(c*x - 1)*x^m/((c^2*m^2 + 4*c^2*m + 3*c^2)*x^2 - m^2 - 4*m - 3), x))/(m^2 + 4
*m + 3)

Giac [F]

\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsec(c*x) + a)*(f*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int {\left (f\,x\right )}^m\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

[In]

int((f*x)^m*(d + e*x^2)*(a + b*acos(1/(c*x))),x)

[Out]

int((f*x)^m*(d + e*x^2)*(a + b*acos(1/(c*x))), x)