3.4.10 \(\int (c e+d e x)^m (a+b \text {ArcSin}(c+d x))^2 \, dx\) [310]

Optimal. Leaf size=183 \[ \frac {(e (c+d x))^{1+m} (a+b \text {ArcSin}(c+d x))^2}{d e (1+m)}-\frac {2 b (e (c+d x))^{2+m} (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};(c+d x)^2\right )}{d e^2 (1+m) (2+m)}+\frac {2 b^2 (e (c+d x))^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};(c+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4889, 4723, 4805} \begin {gather*} \frac {2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac {2 b (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right ) (a+b \text {ArcSin}(c+d x))}{d e^2 (m+1) (m+2)}+\frac {(e (c+d x))^{m+1} (a+b \text {ArcSin}(c+d x))^2}{d e (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(c + d*x))^(1 + m)*(a + b*ArcSin[c + d*x])^2)/(d*e*(1 + m)) - (2*b*(e*(c + d*x))^(2 + m)*(a + b*ArcSin[c +
d*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, (c + d*x)^2])/(d*e^2*(1 + m)*(2 + m)) + (2*b^2*(e*(c + d*x
))^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, (c + d*x)^2])/(d*e^3*(1 + m)*(2
+ m)*(3 + m))

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 -
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -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]

Rule 4889

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int (c e+d e x)^m \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int (e x)^m \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {(e (c+d x))^{1+m} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e (1+m)}-\frac {(2 b) \text {Subst}\left (\int \frac {(e x)^{1+m} \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e (1+m)}\\ &=\frac {(e (c+d x))^{1+m} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e (1+m)}-\frac {2 b (e (c+d x))^{2+m} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};(c+d x)^2\right )}{d e^2 (1+m) (2+m)}+\frac {2 b^2 (e (c+d x))^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};(c+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 151, normalized size = 0.83 \begin {gather*} \frac {(c+d x) (e (c+d x))^m \left ((a+b \text {ArcSin}(c+d x))^2-\frac {2 b (c+d x) (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};(c+d x)^2\right )}{2+m}+\frac {2 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};(c+d x)^2\right )}{(2+m) (3+m)}\right )}{d (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x)*(e*(c + d*x))^m*((a + b*ArcSin[c + d*x])^2 - (2*b*(c + d*x)*(a + b*ArcSin[c + d*x])*Hypergeometric2
F1[1/2, (2 + m)/2, (4 + m)/2, (c + d*x)^2])/(2 + m) + (2*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 3/2 + m/2, 3/2
+ m/2}, {2 + m/2, 5/2 + m/2}, (c + d*x)^2])/((2 + m)*(3 + m))))/(d*(1 + m))

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Maple [F]
time = 1.38, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{m} \left (a +b \arcsin \left (d x +c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

(d*x*e + c*e)^(m + 1)*a^2*e^(-1)/(d*(m + 1)) + ((b^2*d*x*e^m + b^2*c*e^m)*(d*x + c)^m*arctan2(d*x + c, sqrt(d*
x + c + 1)*sqrt(-d*x - c + 1))^2 + (d*m + d)*integrate(2*((b^2*d*x*e^m + b^2*c*e^m)*sqrt(d*x + c + 1)*sqrt(-d*
x - c + 1)*(d*x + c)^m*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + ((a*b*d^2*m + a*b*d^2)*x^2*e^m
+ 2*(a*b*c*d*m + a*b*c*d)*x*e^m + (a*b*c^2 - a*b + (a*b*c^2 - a*b)*m)*e^m)*(d*x + c)^m*arctan2(d*x + c, sqrt(
d*x + c + 1)*sqrt(-d*x - c + 1)))/((d^2*m + d^2)*x^2 + c^2 + (c^2 - 1)*m + 2*(c*d*m + c*d)*x - 1), x))/(d*m +
d)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*(c + d*x))**m*(a + b*asin(c + d*x))**2, 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((d*e*x+c*e)^m*(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^m\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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