3.2.53 \(\int \frac {x^m (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [153]

Optimal. Leaf size=272 \[ \frac {x^{1+m} (a+b \text {ArcSin}(c x))}{d \sqrt {d-c^2 d x^2}}-\frac {m x^{1+m} \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{d (1+m) \sqrt {d-c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1-c^2 x^2} \text {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{d (2+m) \sqrt {d-c^2 d x^2}}+\frac {b c m x^{2+m} \sqrt {1-c^2 x^2} \text {HypergeometricPFQ}\left (\left \{1,1+\frac {m}{2},1+\frac {m}{2}\right \},\left \{\frac {3}{2}+\frac {m}{2},2+\frac {m}{2}\right \},c^2 x^2\right )}{d \left (2+3 m+m^2\right ) \sqrt {d-c^2 d x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4793, 4805, 371} \begin {gather*} \frac {b c m \sqrt {1-c^2 x^2} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{d \left (m^2+3 m+2\right ) \sqrt {d-c^2 d x^2}}-\frac {m \sqrt {1-c^2 x^2} x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{d (m+1) \sqrt {d-c^2 d x^2}}+\frac {x^{m+1} (a+b \text {ArcSin}(c x))}{d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} x^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{d (m+2) \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(a + b*ArcSin[c*x]))/(d*Sqrt[d - c^2*d*x^2]) - (m*x^(1 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*H
ypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(d*(1 + m)*Sqrt[d - c^2*d*x^2]) - (b*c*x^(2 + m)*Sqrt[1
- c^2*x^2]*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, c^2*x^2])/(d*(2 + m)*Sqrt[d - c^2*d*x^2]) + (b*c*m*x^(2
+ m)*Sqrt[1 - c^2*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(d*(2 + 3*m +
m^2)*Sqrt[d - c^2*d*x^2])

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 4793

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
-(f*x)^(m + 1))*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*d*f*(p + 1))), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*c*(n/(2*f*(p + 1)))*Simp[(d + e*
x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; Fre
eQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 1])

Rule 4805

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, c^2*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d +
e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e,
f, m}, x] && EqQ[c^2*d + e, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {m \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx}{d}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {x^{1+m}}{1-c^2 x^2} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1-c^2 x^2} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{d (2+m) \sqrt {d-c^2 d x^2}}-\frac {\left (m \sqrt {1-c^2 x^2}\right ) \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {m x^{1+m} \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{d (1+m) \sqrt {d-c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1-c^2 x^2} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{d (2+m) \sqrt {d-c^2 d x^2}}+\frac {b c m x^{2+m} \sqrt {1-c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{d \left (2+3 m+m^2\right ) \sqrt {d-c^2 d x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.16, size = 207, normalized size = 0.76 \begin {gather*} \frac {x^{1+m} \left (-m (2+m) \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )+(1+m) \left ((2+m) (a+b \text {ArcSin}(c x))-b c x \sqrt {1-c^2 x^2} \text {Hypergeometric2F1}\left (1,1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )\right )+b c m x \sqrt {1-c^2 x^2} \text {HypergeometricPFQ}\left (\left \{1,1+\frac {m}{2},1+\frac {m}{2}\right \},\left \{\frac {3}{2}+\frac {m}{2},2+\frac {m}{2}\right \},c^2 x^2\right )\right )}{d (1+m) (2+m) \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(-(m*(2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2
*x^2]) + (1 + m)*((2 + m)*(a + b*ArcSin[c*x]) - b*c*x*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1, 1 + m/2, 2 + m/2,
c^2*x^2]) + b*c*m*x*Sqrt[1 - c^2*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2]
))/(d*(1 + m)*(2 + m)*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Maple [F]
time = 0.82, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(x^m*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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(x^m*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(3/2),x)

[Out]

Integral(x**m*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

________________________________________________________________________________________

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

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)^(3/2),x)

[Out]

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

________________________________________________________________________________________