3.290 \(\int \frac {a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{11/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac {20 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{189 d e^{11/2}}-\frac {20 b \sqrt {1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac {4 b \sqrt {1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}} \]

[Out]

-2/9*(a+b*arcsin(d*x+c))/d/e/(e*(d*x+c))^(9/2)+20/189*b*EllipticF((e*(d*x+c))^(1/2)/e^(1/2),I)/d/e^(11/2)-4/63
*b*(1-(d*x+c)^2)^(1/2)/d/e^2/(e*(d*x+c))^(7/2)-20/189*b*(1-(d*x+c)^2)^(1/2)/d/e^4/(e*(d*x+c))^(3/2)

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Rubi [A]  time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4805, 4627, 325, 329, 221} \[ -\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}-\frac {20 b \sqrt {1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac {4 b \sqrt {1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}+\frac {20 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{189 d e^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^(11/2),x]

[Out]

(-4*b*Sqrt[1 - (c + d*x)^2])/(63*d*e^2*(e*(c + d*x))^(7/2)) - (20*b*Sqrt[1 - (c + d*x)^2])/(189*d*e^4*(e*(c +
d*x))^(3/2)) - (2*(a + b*ArcSin[c + d*x]))/(9*d*e*(e*(c + d*x))^(9/2)) + (20*b*EllipticF[ArcSin[Sqrt[e*(c + d*
x)]/Sqrt[e]], -1])/(189*d*e^(11/2))

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 325

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 \frac {a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{11/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{(e x)^{11/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{(e x)^{9/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{9 d e}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac {(10 b) \operatorname {Subst}\left (\int \frac {1}{(e x)^{5/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{63 d e^3}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac {20 b \sqrt {1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac {(10 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{189 d e^5}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac {20 b \sqrt {1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac {(20 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{189 d e^6}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac {20 b \sqrt {1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac {20 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{189 d e^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 66, normalized size = 0.47 \[ -\frac {2 \sqrt {e (c+d x)} \left (7 \left (a+b \sin ^{-1}(c+d x)\right )+2 b (c+d x) \, _2F_1\left (-\frac {7}{4},\frac {1}{2};-\frac {3}{4};(c+d x)^2\right )\right )}{63 d e^6 (c+d x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^(11/2),x]

[Out]

(-2*Sqrt[e*(c + d*x)]*(7*(a + b*ArcSin[c + d*x]) + 2*b*(c + d*x)*Hypergeometric2F1[-7/4, 1/2, -3/4, (c + d*x)^
2]))/(63*d*e^6*(c + d*x)^5)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \arcsin \left (d x + c\right ) + a\right )}}{d^{6} e^{6} x^{6} + 6 \, c d^{5} e^{6} x^{5} + 15 \, c^{2} d^{4} e^{6} x^{4} + 20 \, c^{3} d^{3} e^{6} x^{3} + 15 \, c^{4} d^{2} e^{6} x^{2} + 6 \, c^{5} d e^{6} x + c^{6} e^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(11/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*e*x + c*e)*(b*arcsin(d*x + c) + a)/(d^6*e^6*x^6 + 6*c*d^5*e^6*x^5 + 15*c^2*d^4*e^6*x^4 + 20*c^
3*d^3*e^6*x^3 + 15*c^4*d^2*e^6*x^2 + 6*c^5*d*e^6*x + c^6*e^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(11/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)/(d*e*x + c*e)^(11/2), x)

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maple [A]  time = 0.02, size = 203, normalized size = 1.46 \[ \frac {-\frac {2 a}{9 \left (d e x +c e \right )^{\frac {9}{2}}}+2 b \left (-\frac {\arcsin \left (\frac {d e x +c e}{e}\right )}{9 \left (d e x +c e \right )^{\frac {9}{2}}}+\frac {-\frac {2 \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{63 \left (d e x +c e \right )^{\frac {7}{2}}}-\frac {10 \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{189 e^{2} \left (d e x +c e \right )^{\frac {3}{2}}}+\frac {10 \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {\frac {d e x +c e}{e}+1}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )}{189 e^{4} \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(11/2),x)

[Out]

2/d/e*(-1/9*a/(d*e*x+c*e)^(9/2)+b*(-1/9/(d*e*x+c*e)^(9/2)*arcsin((d*e*x+c*e)/e)+2/9/e*(-1/7*(-(d*e*x+c*e)^2/e^
2+1)^(1/2)/(d*e*x+c*e)^(7/2)-5/21/e^2*(-(d*e*x+c*e)^2/e^2+1)^(1/2)/(d*e*x+c*e)^(3/2)+5/21/e^4/(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))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, {\left ({\left (d x + c\right )}^{\frac {9}{2}} b \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + {\left (a d^{4} x^{4} + 4 \, a c d^{3} x^{3} + 6 \, a c^{2} d^{2} x^{2} + {\left (b d^{5} e^{6} x^{4} + 4 \, b c d^{4} e^{6} x^{3} + 6 \, b c^{2} d^{3} e^{6} x^{2} + 4 \, b c^{3} d^{2} e^{6} x + b c^{4} d e^{6}\right )} {\left (d x + c\right )}^{\frac {9}{2}} \int \frac {\sqrt {d x + c + 1} \sqrt {d x + c} \sqrt {-d x - c + 1}}{d^{7} e^{6} x^{7} + 7 \, c d^{6} e^{6} x^{6} + {\left (21 \, c^{2} - 1\right )} d^{5} e^{6} x^{5} + 5 \, {\left (7 \, c^{3} - c\right )} d^{4} e^{6} x^{4} + 5 \, {\left (7 \, c^{4} - 2 \, c^{2}\right )} d^{3} e^{6} x^{3} + {\left (21 \, c^{5} - 10 \, c^{3}\right )} d^{2} e^{6} x^{2} + {\left (7 \, c^{6} - 5 \, c^{4}\right )} d e^{6} x + {\left (c^{7} - c^{5}\right )} e^{6}}\,{d x} + 4 \, a c^{3} d x + a c^{4}\right )} \sqrt {d x + c}\right )}}{9 \, {\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )} {\left (d x + c\right )}^{5} \sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(11/2),x, algorithm="maxima")

[Out]

-2/9*((d*x + c)^(9/2)*b*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + (a*d^4*x^4 + 4*a*c*d^3*x^3 +
6*a*c^2*d^2*x^2 + 9*(b*d^5*e^6*x^4 + 4*b*c*d^4*e^6*x^3 + 6*b*c^2*d^3*e^6*x^2 + 4*b*c^3*d^2*e^6*x + b*c^4*d*e^6
)*(d*x + c)^(9/2)*integrate(1/9*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)/(d^7*e^6*x^7 + 7*c*d^6*e^6*
x^6 + (21*c^2 - 1)*d^5*e^6*x^5 + 5*(7*c^3 - c)*d^4*e^6*x^4 + 5*(7*c^4 - 2*c^2)*d^3*e^6*x^3 + (21*c^5 - 10*c^3)
*d^2*e^6*x^2 + (7*c^6 - 5*c^4)*d*e^6*x + (c^7 - c^5)*e^6), x) + 4*a*c^3*d*x + a*c^4)*sqrt(d*x + c))/((d^5*e^5*
x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)*(d*x + c)^5*sqrt(e))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^(11/2),x)

[Out]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^(11/2), 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((a+b*asin(d*x+c))/(d*e*x+c*e)**(11/2),x)

[Out]

Timed out

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