3.144 \(\int x^m (d-c^2 d x^2)^2 (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=217 \[ \frac {c^4 d^2 x^{m+5} \left (a+b \sin ^{-1}(c x)\right )}{m+5}-\frac {2 c^2 d^2 x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac {d^2 x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac {b c d^2 \left (15 m^2+100 m+149\right ) x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{(m+1) (m+2) (m+3)^2 (m+5)^2}-\frac {b c d^2 \left (m^2+13 m+38\right ) \sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2 (m+5)^2}+\frac {b c^3 d^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2} \]

[Out]

d^2*x^(1+m)*(a+b*arcsin(c*x))/(1+m)-2*c^2*d^2*x^(3+m)*(a+b*arcsin(c*x))/(3+m)+c^4*d^2*x^(5+m)*(a+b*arcsin(c*x)
)/(5+m)-b*c*d^2*(15*m^2+100*m+149)*x^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/(m^2+3*m+2)/(m^2+8*m+15
)^2-b*c*d^2*(m^2+13*m+38)*x^(2+m)*(-c^2*x^2+1)^(1/2)/(3+m)^2/(5+m)^2+b*c^3*d^2*x^(4+m)*(-c^2*x^2+1)^(1/2)/(5+m
)^2

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Rubi [A]  time = 0.31, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {270, 4687, 12, 1267, 459, 364} \[ -\frac {2 c^2 d^2 x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac {c^4 d^2 x^{m+5} \left (a+b \sin ^{-1}(c x)\right )}{m+5}+\frac {d^2 x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac {b c d^2 \left (15 m^2+100 m+149\right ) x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{(m+1) (m+2) (m+3)^2 (m+5)^2}-\frac {b c d^2 \left (m^2+13 m+38\right ) \sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2 (m+5)^2}+\frac {b c^3 d^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]),x]

[Out]

-((b*c*d^2*(38 + 13*m + m^2)*x^(2 + m)*Sqrt[1 - c^2*x^2])/((3 + m)^2*(5 + m)^2)) + (b*c^3*d^2*x^(4 + m)*Sqrt[1
- c^2*x^2])/(5 + m)^2 + (d^2*x^(1 + m)*(a + b*ArcSin[c*x]))/(1 + m) - (2*c^2*d^2*x^(3 + m)*(a + b*ArcSin[c*x]
))/(3 + m) + (c^4*d^2*x^(5 + m)*(a + b*ArcSin[c*x]))/(5 + m) - (b*c*d^2*(149 + 100*m + 15*m^2)*x^(2 + m)*Hyper
geometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((1 + m)*(2 + m)*(3 + m)^2*(5 + m)^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 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 4687

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = I
ntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^m \left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac {c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-(b c) \int \frac {d^2 x^{1+m} \left (\frac {1}{1+m}-\frac {2 c^2 x^2}{3+m}+\frac {c^4 x^4}{5+m}\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac {c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-\left (b c d^2\right ) \int \frac {x^{1+m} \left (\frac {1}{1+m}-\frac {2 c^2 x^2}{3+m}+\frac {c^4 x^4}{5+m}\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b c^3 d^2 x^{4+m} \sqrt {1-c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac {c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}+\frac {\left (b d^2\right ) \int \frac {x^{1+m} \left (-\frac {c^2 (5+m)}{1+m}+\frac {c^4 \left (38+13 m+m^2\right ) x^2}{(3+m) (5+m)}\right )}{\sqrt {1-c^2 x^2}} \, dx}{c (5+m)}\\ &=-\frac {b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2 (5+m)^2}+\frac {b c^3 d^2 x^{4+m} \sqrt {1-c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac {c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-\frac {\left (b c d^2 \left (149+100 m+15 m^2\right )\right ) \int \frac {x^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{(1+m) (3+m)^2 (5+m)^2}\\ &=-\frac {b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2 (5+m)^2}+\frac {b c^3 d^2 x^{4+m} \sqrt {1-c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac {c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-\frac {b c d^2 \left (149+100 m+15 m^2\right ) x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{(1+m) (2+m) (3+m)^2 (5+m)^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 187, normalized size = 0.86 \[ \frac {x^{m+1} \left (-\frac {4 d^2 \left ((m+2) \left (m \left (c^2 x^2-1\right )+c^2 x^2-3\right ) \left (a+b \sin ^{-1}(c x)\right )+b c (m+1) x \, _2F_1\left (-\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )+2 b c x \, _2F_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )\right )}{(m+1) (m+2) (m+3)}+\left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 x \, _2F_1\left (-\frac {3}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )}{m+2}\right )}{m+5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]),x]

[Out]

(x^(1 + m)*((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]) - (b*c*d^2*x*Hypergeometric2F1[-3/2, 1 + m/2, 2 + m/2, c^2*x
^2])/(2 + m) - (4*d^2*((2 + m)*(-3 + c^2*x^2 + m*(-1 + c^2*x^2))*(a + b*ArcSin[c*x]) + b*c*(1 + m)*x*Hypergeom
etric2F1[-1/2, 1 + m/2, 2 + m/2, c^2*x^2] + 2*b*c*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, c^2*x^2]))/((1 +
m)*(2 + m)*(3 + m))))/(5 + m)

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fricas [F]  time = 1.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-c^2*d*x^2+d)^2*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*x^m
, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-c^2*d*x^2+d)^2*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((c^2*d*x^2 - d)^2*(b*arcsin(c*x) + a)*x^m, x)

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maple [F]  time = 7.35, size = 0, normalized size = 0.00 \[ \int x^{m} \left (-c^{2} d \,x^{2}+d \right )^{2} \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(-c^2*d*x^2+d)^2*(a+b*arcsin(c*x)),x)

[Out]

int(x^m*(-c^2*d*x^2+d)^2*(a+b*arcsin(c*x)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-c^2*d*x^2+d)^2*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

a*c^4*d^2*x^(m + 5)/(m + 5) - 2*a*c^2*d^2*x^(m + 3)/(m + 3) + a*d^2*x^(m + 1)/(m + 1) + (((b*c^4*d^2*m^2 + 4*b
*c^4*d^2*m + 3*b*c^4*d^2)*x^5 - 2*(b*c^2*d^2*m^2 + 6*b*c^2*d^2*m + 5*b*c^2*d^2)*x^3 + (b*d^2*m^2 + 8*b*d^2*m +
15*b*d^2)*x)*x^m*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (m^3 + 9*m^2 + 23*m + 15)*integrate(-((b*c^5*d^
2*m^2 + 4*b*c^5*d^2*m + 3*b*c^5*d^2)*x^5 - 2*(b*c^3*d^2*m^2 + 6*b*c^3*d^2*m + 5*b*c^3*d^2)*x^3 + (b*c*d^2*m^2
+ 8*b*c*d^2*m + 15*b*c*d^2)*x)*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 \[ \int x^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a + b*asin(c*x))*(d - c^2*d*x^2)^2,x)

[Out]

int(x^m*(a + b*asin(c*x))*(d - c^2*d*x^2)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} \left (\int a x^{m}\, dx + \int b x^{m} \operatorname {asin}{\left (c x \right )}\, dx + \int \left (- 2 a c^{2} x^{2} x^{m}\right )\, dx + \int a c^{4} x^{4} x^{m}\, dx + \int \left (- 2 b c^{2} x^{2} x^{m} \operatorname {asin}{\left (c x \right )}\right )\, dx + \int b c^{4} x^{4} x^{m} \operatorname {asin}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(-c**2*d*x**2+d)**2*(a+b*asin(c*x)),x)

[Out]

d**2*(Integral(a*x**m, x) + Integral(b*x**m*asin(c*x), x) + Integral(-2*a*c**2*x**2*x**m, x) + Integral(a*c**4
*x**4*x**m, x) + Integral(-2*b*c**2*x**2*x**m*asin(c*x), x) + Integral(b*c**4*x**4*x**m*asin(c*x), x))

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