3.7.54 \(\int (f x)^m (d+e x^2)^3 (a+b \text {ArcSin}(c x)) \, dx\) [654]

Optimal. Leaf size=484 \[ \frac {b e \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right ) (f x)^{2+m} \sqrt {1-c^2 x^2}}{c^5 f^2 (3+m)^2 (5+m)^2 (7+m)^2}+\frac {b e^2 \left (3 c^2 d (7+m)^2+e \left (30+11 m+m^2\right )\right ) (f x)^{4+m} \sqrt {1-c^2 x^2}}{c^3 f^4 (5+m)^2 (7+m)^2}+\frac {b e^3 (f x)^{6+m} \sqrt {1-c^2 x^2}}{c f^6 (7+m)^2}+\frac {d^3 (f x)^{1+m} (a+b \text {ArcSin}(c x))}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} (a+b \text {ArcSin}(c x))}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} (a+b \text {ArcSin}(c x))}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} (a+b \text {ArcSin}(c x))}{f^7 (7+m)}-\frac {b \left (\frac {c^6 d^3 (3+m) (5+m) (7+m)}{1+m}+\frac {e (2+m) \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right )}{(3+m) (5+m) (7+m)}\right ) (f x)^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{c^5 f^2 (2+m) (3+m) (5+m) (7+m)} \]

[Out]

d^3*(f*x)^(1+m)*(a+b*arcsin(c*x))/f/(1+m)+3*d^2*e*(f*x)^(3+m)*(a+b*arcsin(c*x))/f^3/(3+m)+3*d*e^2*(f*x)^(5+m)*
(a+b*arcsin(c*x))/f^5/(5+m)+e^3*(f*x)^(7+m)*(a+b*arcsin(c*x))/f^7/(7+m)-b*(c^6*d^3*(3+m)*(5+m)*(7+m)/(1+m)+e*(
2+m)*(3*c^2*d*e*(7+m)^2*(m^2+7*m+12)+3*c^4*d^2*(m^2+12*m+35)^2+e^2*(m^4+18*m^3+119*m^2+342*m+360))/(m^3+15*m^2
+71*m+105))*(f*x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/c^5/f^2/(2+m)/(3+m)/(5+m)/(7+m)+b*e*(3*c^2
*d*e*(7+m)^2*(m^2+7*m+12)+3*c^4*d^2*(m^2+12*m+35)^2+e^2*(m^4+18*m^3+119*m^2+342*m+360))*(f*x)^(2+m)*(-c^2*x^2+
1)^(1/2)/c^5/f^2/(3+m)^2/(5+m)^2/(7+m)^2+b*e^2*(3*c^2*d*(7+m)^2+e*(m^2+11*m+30))*(f*x)^(4+m)*(-c^2*x^2+1)^(1/2
)/c^3/f^4/(5+m)^2/(7+m)^2+b*e^3*(f*x)^(6+m)*(-c^2*x^2+1)^(1/2)/c/f^6/(7+m)^2

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Rubi [A]
time = 1.60, antiderivative size = 455, normalized size of antiderivative = 0.94, 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 = {276, 4815, 12, 1823, 1281, 470, 371} \begin {gather*} \frac {d^3 (f x)^{m+1} (a+b \text {ArcSin}(c x))}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} (a+b \text {ArcSin}(c x))}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} (a+b \text {ArcSin}(c x))}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} (a+b \text {ArcSin}(c x))}{f^7 (m+7)}+\frac {b e^3 \sqrt {1-c^2 x^2} (f x)^{m+6}}{c f^6 (m+7)^2}+\frac {b e^2 \sqrt {1-c^2 x^2} (f x)^{m+4} \left (3 c^2 d (m+7)^2+e \left (m^2+11 m+30\right )\right )}{c^3 f^4 (m+5)^2 (m+7)^2}-\frac {b c (f x)^{m+2} \left (\frac {e \left (3 c^4 d^2 \left (m^2+12 m+35\right )^2+3 c^2 d e (m+7)^2 \left (m^2+7 m+12\right )+e^2 \left (m^4+18 m^3+119 m^2+342 m+360\right )\right )}{c^6 (m+3)^2 (m+5)^2 (m+7)^2}+\frac {d^3}{m^2+3 m+2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2}+\frac {b e \sqrt {1-c^2 x^2} (f x)^{m+2} \left (3 c^4 d^2 \left (m^2+12 m+35\right )^2+3 c^2 d e (m+7)^2 \left (m^2+7 m+12\right )+e^2 \left (m^4+18 m^3+119 m^2+342 m+360\right )\right )}{c^5 f^2 (m+3)^2 (m+5)^2 (m+7)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*(d + e*x^2)^3*(a + b*ArcSin[c*x]),x]

[Out]

(b*e*(3*c^2*d*e*(7 + m)^2*(12 + 7*m + m^2) + 3*c^4*d^2*(35 + 12*m + m^2)^2 + e^2*(360 + 342*m + 119*m^2 + 18*m
^3 + m^4))*(f*x)^(2 + m)*Sqrt[1 - c^2*x^2])/(c^5*f^2*(3 + m)^2*(5 + m)^2*(7 + m)^2) + (b*e^2*(3*c^2*d*(7 + m)^
2 + e*(30 + 11*m + m^2))*(f*x)^(4 + m)*Sqrt[1 - c^2*x^2])/(c^3*f^4*(5 + m)^2*(7 + m)^2) + (b*e^3*(f*x)^(6 + m)
*Sqrt[1 - c^2*x^2])/(c*f^6*(7 + m)^2) + (d^3*(f*x)^(1 + m)*(a + b*ArcSin[c*x]))/(f*(1 + m)) + (3*d^2*e*(f*x)^(
3 + m)*(a + b*ArcSin[c*x]))/(f^3*(3 + m)) + (3*d*e^2*(f*x)^(5 + m)*(a + b*ArcSin[c*x]))/(f^5*(5 + m)) + (e^3*(
f*x)^(7 + m)*(a + b*ArcSin[c*x]))/(f^7*(7 + m)) - (b*c*(d^3/(2 + 3*m + m^2) + (e*(3*c^2*d*e*(7 + m)^2*(12 + 7*
m + m^2) + 3*c^4*d^2*(35 + 12*m + m^2)^2 + e^2*(360 + 342*m + 119*m^2 + 18*m^3 + m^4)))/(c^6*(3 + m)^2*(5 + m)
^2*(7 + m)^2))*(f*x)^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/f^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 276

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 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 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 1281

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 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 4815

Int[((a_.) + ArcSin[(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*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] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rubi steps

\begin {align*} \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {d^3 (f x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \sin ^{-1}(c x)\right )}{f^7 (7+m)}-(b c) \int \frac {(f x)^{1+m} \left (\frac {d^3}{1+m}+\frac {3 d^2 e x^2}{3+m}+\frac {3 d e^2 x^4}{5+m}+\frac {e^3 x^6}{7+m}\right )}{f \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d^3 (f x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \sin ^{-1}(c x)\right )}{f^7 (7+m)}-\frac {(b c) \int \frac {(f x)^{1+m} \left (\frac {d^3}{1+m}+\frac {3 d^2 e x^2}{3+m}+\frac {3 d e^2 x^4}{5+m}+\frac {e^3 x^6}{7+m}\right )}{\sqrt {1-c^2 x^2}} \, dx}{f}\\ &=\frac {b e^3 (f x)^{6+m} \sqrt {1-c^2 x^2}}{c f^6 (7+m)^2}+\frac {d^3 (f x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \sin ^{-1}(c x)\right )}{f^7 (7+m)}+\frac {b \int \frac {(f x)^{1+m} \left (-\frac {c^2 d^3 (7+m)}{1+m}-\frac {3 c^2 d^2 e (7+m) x^2}{3+m}-\frac {e^2 \left (3 c^2 d (7+m)^2+e \left (30+11 m+m^2\right )\right ) x^4}{(5+m) (7+m)}\right )}{\sqrt {1-c^2 x^2}} \, dx}{c f (7+m)}\\ &=\frac {b e^2 \left (3 c^2 d (7+m)^2+e \left (30+11 m+m^2\right )\right ) (f x)^{4+m} \sqrt {1-c^2 x^2}}{c^3 f^4 (5+m)^2 (7+m)^2}+\frac {b e^3 (f x)^{6+m} \sqrt {1-c^2 x^2}}{c f^6 (7+m)^2}+\frac {d^3 (f x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \sin ^{-1}(c x)\right )}{f^7 (7+m)}-\frac {b \int \frac {(f x)^{1+m} \left (\frac {c^4 d^3 (5+m) (7+m)}{1+m}+\frac {e \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right ) x^2}{(3+m) (5+m) (7+m)}\right )}{\sqrt {1-c^2 x^2}} \, dx}{c^3 f (5+m) (7+m)}\\ &=\frac {b e \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right ) (f x)^{2+m} \sqrt {1-c^2 x^2}}{c^5 f^2 (3+m)^2 (5+m)^2 (7+m)^2}+\frac {b e^2 \left (3 c^2 d (7+m)^2+e \left (30+11 m+m^2\right )\right ) (f x)^{4+m} \sqrt {1-c^2 x^2}}{c^3 f^4 (5+m)^2 (7+m)^2}+\frac {b e^3 (f x)^{6+m} \sqrt {1-c^2 x^2}}{c f^6 (7+m)^2}+\frac {d^3 (f x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \sin ^{-1}(c x)\right )}{f^7 (7+m)}-\frac {\left (b \left (\frac {c^6 d^3}{1+m}+\frac {e (2+m) \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right )}{(3+m)^2 (5+m)^2 (7+m)^2}\right )\right ) \int \frac {(f x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{c^5 f}\\ &=\frac {b e \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right ) (f x)^{2+m} \sqrt {1-c^2 x^2}}{c^5 f^2 (3+m)^2 (5+m)^2 (7+m)^2}+\frac {b e^2 \left (3 c^2 d (7+m)^2+e \left (30+11 m+m^2\right )\right ) (f x)^{4+m} \sqrt {1-c^2 x^2}}{c^3 f^4 (5+m)^2 (7+m)^2}+\frac {b e^3 (f x)^{6+m} \sqrt {1-c^2 x^2}}{c f^6 (7+m)^2}+\frac {d^3 (f x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \sin ^{-1}(c x)\right )}{f^7 (7+m)}-\frac {b \left (\frac {c^6 d^3}{1+m}+\frac {e (2+m) \left (3 c^2 d e (7+m)^2 \left (12+7 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )^2+e^2 \left (360+342 m+119 m^2+18 m^3+m^4\right )\right )}{(3+m)^2 (5+m)^2 (7+m)^2}\right ) (f x)^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{c^5 f^2 (2+m)}\\ \end {align*}

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Mathematica [F]
time = 5.19, size = 0, normalized size = 0.00 \begin {gather*} \int (f x)^m \left (d+e x^2\right )^3 (a+b \text {ArcSin}(c x)) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(f*x)^m*(d + e*x^2)^3*(a + b*ArcSin[c*x]),x]

[Out]

Integrate[(f*x)^m*(d + e*x^2)^3*(a + b*ArcSin[c*x]), x]

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Maple [F]
time = 18.97, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{3} \left (a +b \arcsin \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(e*x^2+d)^3*(a+b*arcsin(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)^3*(a+b*arcsin(c*x)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)^3*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

a*f^m*x^7*e^(m*log(x) + 3)/(m + 7) + 3*a*d*f^m*x^5*e^(m*log(x) + 2)/(m + 5) + 3*a*d^2*f^m*x^3*e^(m*log(x) + 1)
/(m + 3) + (f*x)^(m + 1)*a*d^3/(f*(m + 1)) + (((b*f^m*m^3*e^3 + 9*b*f^m*m^2*e^3 + 23*b*f^m*m*e^3 + 15*b*f^m*e^
3)*x^7 + 3*(b*d*f^m*m^3*e^2 + 11*b*d*f^m*m^2*e^2 + 31*b*d*f^m*m*e^2 + 21*b*d*f^m*e^2)*x^5 + 3*(b*d^2*f^m*m^3*e
+ 13*b*d^2*f^m*m^2*e + 47*b*d^2*f^m*m*e + 35*b*d^2*f^m*e)*x^3 + (b*d^3*f^m*m^3 + 15*b*d^3*f^m*m^2 + 71*b*d^3*
f^m*m + 105*b*d^3*f^m)*x)*x^m*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (m^4 + 16*m^3 + 86*m^2 + 176*m + 10
5)*integrate(-((b*c*f^m*m^3*e^3 + 9*b*c*f^m*m^2*e^3 + 23*b*c*f^m*m*e^3 + 15*b*c*f^m*e^3)*x^7 + 3*(b*c*d*f^m*m^
3*e^2 + 11*b*c*d*f^m*m^2*e^2 + 31*b*c*d*f^m*m*e^2 + 21*b*c*d*f^m*e^2)*x^5 + 3*(b*c*d^2*f^m*m^3*e + 13*b*c*d^2*
f^m*m^2*e + 47*b*c*d^2*f^m*m*e + 35*b*c*d^2*f^m*e)*x^3 + (b*c*d^3*f^m*m^3 + 15*b*c*d^3*f^m*m^2 + 71*b*c*d^3*f^
m*m + 105*b*c*d^3*f^m)*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^m/(m^4 + 16*m^3 - (c^2*m^4 + 16*c^2*m^3 + 86*c^2*m^2
+ 176*c^2*m + 105*c^2)*x^2 + 86*m^2 + 176*m + 105), x))/(m^4 + 16*m^3 + 86*m^2 + 176*m + 105)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)^3*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((a*x^6*e^3 + 3*a*d*x^4*e^2 + 3*a*d^2*x^2*e + a*d^3 + (b*x^6*e^3 + 3*b*d*x^4*e^2 + 3*b*d^2*x^2*e + b*d
^3)*arcsin(c*x))*(f*x)^m, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (f x\right )^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(e*x**2+d)**3*(a+b*asin(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*asin(c*x))*(d + e*x**2)**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)^3*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^3*(b*arcsin(c*x) + a)*(f*x)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2)^3,x)

[Out]

int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2)^3, x)

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