3.281 \(\int (c e+d e x)^{7/2} (a+b \sin ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=156 \[ \frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^3 \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{135 d \sqrt {c+d x}}+\frac {28 b e^2 \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}}{405 d}+\frac {4 b \sqrt {1-(c+d x)^2} (e (c+d x))^{7/2}}{81 d} \]

[Out]

2/9*(e*(d*x+c))^(9/2)*(a+b*arcsin(d*x+c))/d/e+28/135*b*e^3*EllipticE(1/2*(-d*x-c+1)^(1/2)*2^(1/2),2^(1/2))*(e*
(d*x+c))^(1/2)/d/(d*x+c)^(1/2)+28/405*b*e^2*(e*(d*x+c))^(3/2)*(1-(d*x+c)^2)^(1/2)/d+4/81*b*(e*(d*x+c))^(7/2)*(
1-(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4805, 4627, 321, 320, 318, 424} \[ \frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^2 \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}}{405 d}+\frac {28 b e^3 \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{135 d \sqrt {c+d x}}+\frac {4 b \sqrt {1-(c+d x)^2} (e (c+d x))^{7/2}}{81 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(7/2)*(a + b*ArcSin[c + d*x]),x]

[Out]

(28*b*e^2*(e*(c + d*x))^(3/2)*Sqrt[1 - (c + d*x)^2])/(405*d) + (4*b*(e*(c + d*x))^(7/2)*Sqrt[1 - (c + d*x)^2])
/(81*d) + (2*(e*(c + d*x))^(9/2)*(a + b*ArcSin[c + d*x]))/(9*d*e) + (28*b*e^3*Sqrt[e*(c + d*x)]*EllipticE[ArcS
in[Sqrt[1 - c - d*x]/Sqrt[2]], 2])/(135*d*Sqrt[c + d*x])

Rule 318

Int[Sqrt[x_]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[-2/(Sqrt[a]*(-(b/a))^(3/4)), Subst[Int[Sqrt[1 - 2*x^
2]/Sqrt[1 - x^2], x], x, Sqrt[1 - Sqrt[-(b/a)]*x]/Sqrt[2]], x] /; FreeQ[{a, b}, x] && GtQ[-(b/a), 0] && GtQ[a,
0]

Rule 320

Int[Sqrt[(c_)*(x_)]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[c*x]/Sqrt[x], Int[Sqrt[x]/Sqrt[a + b*x^2
], x], x] /; FreeQ[{a, b, c}, x] && GtQ[-(b/a), 0]

Rule 321

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 4627

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_) + (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)^{7/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{7/2} \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(e x)^{9/2}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{9 d e}\\ &=\frac {4 b (e (c+d x))^{7/2} \sqrt {1-(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}-\frac {(14 b e) \operatorname {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{81 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{405 d}+\frac {4 b (e (c+d x))^{7/2} \sqrt {1-(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (14 b e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{405 d}+\frac {4 b (e (c+d x))^{7/2} \sqrt {1-(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (14 b e^3 \sqrt {e (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{135 d \sqrt {c+d x}}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{405 d}+\frac {4 b (e (c+d x))^{7/2} \sqrt {1-(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}+\frac {\left (28 b e^3 \sqrt {e (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^2}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {1-c-d x}}{\sqrt {2}}\right )}{135 d \sqrt {c+d x}}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{405 d}+\frac {4 b (e (c+d x))^{7/2} \sqrt {1-(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^3 \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {1-c-d x}}{\sqrt {2}}\right )\right |2\right )}{135 d \sqrt {c+d x}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 115, normalized size = 0.74 \[ \frac {2 (e (c+d x))^{7/2} \left (45 a (c+d x)^3-14 b \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right )+10 b \sqrt {1-(c+d x)^2} (c+d x)^2+14 b \sqrt {1-(c+d x)^2}+45 b (c+d x)^3 \sin ^{-1}(c+d x)\right )}{405 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(7/2)*(a + b*ArcSin[c + d*x]),x]

[Out]

(2*(e*(c + d*x))^(7/2)*(45*a*(c + d*x)^3 + 14*b*Sqrt[1 - (c + d*x)^2] + 10*b*(c + d*x)^2*Sqrt[1 - (c + d*x)^2]
+ 45*b*(c + d*x)^3*ArcSin[c + d*x] - 14*b*Hypergeometric2F1[1/2, 3/4, 7/4, (c + d*x)^2]))/(405*d*(c + d*x)^2)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a d^{3} e^{3} x^{3} + 3 \, a c d^{2} e^{3} x^{2} + 3 \, a c^{2} d e^{3} x + a c^{3} e^{3} + {\left (b d^{3} e^{3} x^{3} + 3 \, b c d^{2} e^{3} x^{2} + 3 \, b c^{2} d e^{3} x + b c^{3} e^{3}\right )} \arcsin \left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c)),x, algorithm="fricas")

[Out]

integral((a*d^3*e^3*x^3 + 3*a*c*d^2*e^3*x^2 + 3*a*c^2*d*e^3*x + a*c^3*e^3 + (b*d^3*e^3*x^3 + 3*b*c*d^2*e^3*x^2
+ 3*b*c^2*d*e^3*x + b*c^3*e^3)*arcsin(d*x + c))*sqrt(d*e*x + c*e), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )}^{\frac {7}{2}} {\left (b \arcsin \left (d x + c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c)),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(7/2)*(b*arcsin(d*x + c) + a), x)

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maple [C]  time = 0.04, size = 228, normalized size = 1.46 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {9}{2}} a}{9}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {9}{2}} \arcsin \left (\frac {d e x +c e}{e}\right )}{9}-\frac {2 \left (-\frac {e^{2} \left (d e x +c e \right )^{\frac {7}{2}} \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{9}-\frac {7 e^{4} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{45}-\frac {7 e^{5} \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {\frac {d e x +c e}{e}+1}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )\right )}{15 \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{9 e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c)),x)

[Out]

2/d/e*(1/9*(d*e*x+c*e)^(9/2)*a+b*(1/9*(d*e*x+c*e)^(9/2)*arcsin((d*e*x+c*e)/e)-2/9/e*(-1/9*e^2*(d*e*x+c*e)^(7/2
)*(-(d*e*x+c*e)^2/e^2+1)^(1/2)-7/45*e^4*(d*e*x+c*e)^(3/2)*(-(d*e*x+c*e)^2/e^2+1)^(1/2)-7/15*e^5/(1/e)^(1/2)*(1
-(d*e*x+c*e)/e)^(1/2)*((d*e*x+c*e)/e+1)^(1/2)/(-(d*e*x+c*e)^2/e^2+1)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(1/e)^
(1/2),I)-EllipticE((d*e*x+c*e)^(1/2)*(1/e)^(1/2),I)))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c)),x, algorithm="maxima")

[Out]

1/9*(2*(b*d^4*e^3*x^4 + 4*b*c*d^3*e^3*x^3 + 6*b*c^2*d^2*e^3*x^2 + 4*b*c^3*d*e^3*x + b*c^4*e^3)*sqrt(d*x + c)*s
qrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + (2*(d*x + c)^(9/2)*a*e^3 + 9*d*integrate(2/9*(
b*d^4*e^3*x^4 + 4*b*c*d^3*e^3*x^3 + 6*b*c^2*d^2*e^3*x^2 + 4*b*c^3*d*e^3*x + b*c^4*e^3)*sqrt(d*x + c + 1)*sqrt(
d*x + c)*sqrt(-d*x - c + 1)/(d^2*x^2 + 2*c*d*x + c^2 - 1), x))*sqrt(e))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,e+d\,e\,x\right )}^{7/2}\,\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^(7/2)*(a + b*asin(c + d*x)),x)

[Out]

int((c*e + d*e*x)^(7/2)*(a + b*asin(c + d*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(7/2)*(a+b*asin(d*x+c)),x)

[Out]

Timed out

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