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

Optimal. Leaf size=374 \[ \frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}-\frac {b e^2 x \sqrt {c^2 x^2-1} (f x)^{m+3}}{c f^3 (m+4) (m+5) \sqrt {c^2 x^2}}-\frac {b e x \sqrt {c^2 x^2-1} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^3 f (m+2) (m+3) (m+4) (m+5) \sqrt {c^2 x^2}}-\frac {b x \sqrt {1-c^2 x^2} (f x)^{m+1} \left (c^4 d^2 (m+2) (m+3) (m+4) (m+5)+e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{c^3 f (m+1)^2 (m+2) (m+3) (m+4) (m+5) \sqrt {c^2 x^2} \sqrt {c^2 x^2-1}} \]

[Out]

d^2*(f*x)^(1+m)*(a+b*arcsec(c*x))/f/(1+m)+2*d*e*(f*x)^(3+m)*(a+b*arcsec(c*x))/f^3/(3+m)+e^2*(f*x)^(5+m)*(a+b*a
rcsec(c*x))/f^5/(5+m)-b*(c^4*d^2*(2+m)*(3+m)*(4+m)*(5+m)+e*(1+m)^2*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20)))*x*(f*x)^(
1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)*(-c^2*x^2+1)^(1/2)/c^3/f/(1+m)^2/(2+m)/(3+m)/(4+m)/(5+m)/
(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)-b*e*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20))*x*(f*x)^(1+m)*(c^2*x^2-1)^(1/2)/c^3/f/(
4+m)/(5+m)/(m^2+5*m+6)/(c^2*x^2)^(1/2)-b*e^2*x*(f*x)^(3+m)*(c^2*x^2-1)^(1/2)/c/f^3/(4+m)/(5+m)/(c^2*x^2)^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 355, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {270, 5238, 12, 1267, 459, 365, 364} \[ \frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}-\frac {b c x \sqrt {1-c^2 x^2} (f x)^{m+1} \left (\frac {e \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 (m+2) (m+3) (m+4) (m+5)}+\frac {d^2}{(m+1)^2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{f \sqrt {c^2 x^2} \sqrt {c^2 x^2-1}}-\frac {b e x \sqrt {c^2 x^2-1} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^3 f (m+2) (m+3) (m+4) (m+5) \sqrt {c^2 x^2}}-\frac {b e^2 x \sqrt {c^2 x^2-1} (f x)^{m+3}}{c f^3 (m+4) (m+5) \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*(e*(3 + m)^2 + 2*c^2*d*(20 + 9*m + m^2))*x*(f*x)^(1 + m)*Sqrt[-1 + c^2*x^2])/(c^3*f*(2 + m)*(3 + m)*(4
+ m)*(5 + m)*Sqrt[c^2*x^2])) - (b*e^2*x*(f*x)^(3 + m)*Sqrt[-1 + c^2*x^2])/(c*f^3*(4 + m)*(5 + m)*Sqrt[c^2*x^2]
) + (d^2*(f*x)^(1 + m)*(a + b*ArcSec[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m)*(a + b*ArcSec[c*x]))/(f^3*(3 +
m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcSec[c*x]))/(f^5*(5 + m)) - (b*c*(d^2/(1 + m)^2 + (e*(e*(3 + m)^2 + 2*c^2*d*
(20 + 9*m + m^2)))/(c^4*(2 + m)*(3 + m)*(4 + m)*(5 + 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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

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 459

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 1267

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[(c^p*(f*x)^(m + 4*p - 1)*(d + e*x^2)^(q + 1))/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1)), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]

Rule 5238

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*} \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{(1+m) (3+m) (5+m) \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{\sqrt {-1+c^2 x^2}} \, dx}{\left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2}}\\ &=-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {(b x) \int \frac {(f x)^m \left (c^2 d^2 (3+m) (4+m) (5+m)+e (1+m) \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x^2\right )}{\sqrt {-1+c^2 x^2}} \, dx}{c (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2}}\\ &=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}--\frac {\left (b \left (-c^4 d^2 (2+m) (3+m) (4+m) (5+m)-e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x\right ) \int \frac {(f x)^m}{\sqrt {-1+c^2 x^2}} \, dx}{c^3 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2}}\\ &=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}--\frac {\left (b \left (-c^4 d^2 (2+m) (3+m) (4+m) (5+m)-e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x \sqrt {1-c^2 x^2}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}} \, dx}{c^3 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}}\\ &=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x (f x)^{1+m} \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{c^3 f (1+m) (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}}\\ \end {align*}

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Mathematica [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} + {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \operatorname {arcsec}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arcsec(c*x))*(f*x)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 7.52, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e^{2} f^{m} x^{5} x^{m}}{m + 5} + \frac {2 \, a d e f^{m} x^{3} x^{m}}{m + 3} + \frac {\left (f x\right )^{m + 1} a d^{2}}{f {\left (m + 1\right )}} + \frac {{\left ({\left (b e^{2} f^{m} m^{2} + 4 \, b e^{2} f^{m} m + 3 \, b e^{2} f^{m}\right )} x^{5} + 2 \, {\left (b d e f^{m} m^{2} + 6 \, b d e f^{m} m + 5 \, b d e f^{m}\right )} x^{3} + {\left (b d^{2} f^{m} m^{2} + 8 \, b d^{2} f^{m} m + 15 \, b d^{2} f^{m}\right )} x\right )} x^{m} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) + {\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} \int \frac {{\left (b d^{2} f^{m} m^{2} + 8 \, b d^{2} f^{m} m + {\left (b e^{2} f^{m} m^{2} + 4 \, b e^{2} f^{m} m + 3 \, b e^{2} f^{m}\right )} x^{4} + 15 \, b d^{2} f^{m} + 2 \, {\left (b d e f^{m} m^{2} + 6 \, b d e f^{m} m + 5 \, b d e f^{m}\right )} x^{2}\right )} \sqrt {c x + 1} \sqrt {c x - 1} x^{m}}{m^{3} - {\left (c^{2} m^{3} + 9 \, c^{2} m^{2} + 23 \, c^{2} m + 15 \, c^{2}\right )} x^{2} + 9 \, m^{2} + 23 \, m + 15}\,{d x}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e^2*f^m*x^5*x^m/(m + 5) + 2*a*d*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d^2/(f*(m + 1)) + (((b*e^2*f^m*m^2 +
4*b*e^2*f^m*m + 3*b*e^2*f^m)*x^5 + 2*(b*d*e*f^m*m^2 + 6*b*d*e*f^m*m + 5*b*d*e*f^m)*x^3 + (b*d^2*f^m*m^2 + 8*b
*d^2*f^m*m + 15*b*d^2*f^m)*x)*x^m*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - (m^3 + 9*m^2 + 23*m + 15)*integrate(-(
b*d^2*f^m*m^2 + 8*b*d^2*f^m*m + (b*e^2*f^m*m^2 + 4*b*e^2*f^m*m + 3*b*e^2*f^m)*x^4 + 15*b*d^2*f^m + 2*(b*d*e*f^
m*m^2 + 6*b*d*e*f^m*m + 5*b*d*e*f^m)*x^2)*sqrt(c*x + 1)*sqrt(c*x - 1)*x^m/(m^3 - (c^2*m^3 + 9*c^2*m^2 + 23*c^2
*m + 15*c^2)*x^2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9*m^2 + 23*m + 15)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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