3.299 \(\int \frac {(a+b \sin ^{-1}(c+d x))^2}{(c e+d e x)^{9/2}} \, dx\)
Optimal. Leaf size=130 \[ -\frac {16 b^2 \, _3F_2\left (-\frac {3}{4},-\frac {3}{4},1;-\frac {1}{4},\frac {1}{4};(c+d x)^2\right )}{105 d e^3 (e (c+d x))^{3/2}}-\frac {8 b \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{35 d e^2 (e (c+d x))^{5/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{7 d e (e (c+d x))^{7/2}} \]
[Out]
-2/7*(a+b*arcsin(d*x+c))^2/d/e/(e*(d*x+c))^(7/2)-8/35*b*(a+b*arcsin(d*x+c))*hypergeom([-5/4, 1/2],[-1/4],(d*x+
c)^2)/d/e^2/(e*(d*x+c))^(5/2)-16/105*b^2*HypergeometricPFQ([-3/4, -3/4, 1],[-1/4, 1/4],(d*x+c)^2)/d/e^3/(e*(d*
x+c))^(3/2)
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Rubi [A] time = 0.22, antiderivative size = 130, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used =
{4805, 4627, 4711} \[ -\frac {16 b^2 \, _3F_2\left (-\frac {3}{4},-\frac {3}{4},1;-\frac {1}{4},\frac {1}{4};(c+d x)^2\right )}{105 d e^3 (e (c+d x))^{3/2}}-\frac {8 b \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{35 d e^2 (e (c+d x))^{5/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{7 d e (e (c+d x))^{7/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^(9/2),x]
[Out]
(-2*(a + b*ArcSin[c + d*x])^2)/(7*d*e*(e*(c + d*x))^(7/2)) - (8*b*(a + b*ArcSin[c + d*x])*Hypergeometric2F1[-5
/4, 1/2, -1/4, (c + d*x)^2])/(35*d*e^2*(e*(c + d*x))^(5/2)) - (16*b^2*HypergeometricPFQ[{-3/4, -3/4, 1}, {-1/4
, 1/4}, (c + d*x)^2])/(105*d*e^3*(e*(c + d*x))^(3/2))
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 4711
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(Sqrt[d]*f*(m + 1)), x] -
Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[d]*f^2*
(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && !IntegerQ[m]
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 {\left (a+b \sin ^{-1}(c+d x)\right )^2}{(c e+d e x)^{9/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{(e x)^{9/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{7 d e (e (c+d x))^{7/2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{(e x)^{7/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{7 d e}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{7 d e (e (c+d x))^{7/2}}-\frac {8 b \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};(c+d x)^2\right )}{35 d e^2 (e (c+d x))^{5/2}}-\frac {16 b^2 \, _3F_2\left (-\frac {3}{4},-\frac {3}{4},1;-\frac {1}{4},\frac {1}{4};(c+d x)^2\right )}{105 d e^3 (e (c+d x))^{3/2}}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 114, normalized size = 0.88 \[ -\frac {2 \sqrt {e (c+d x)} \left (8 b^2 (c+d x)^2 \, _3F_2\left (-\frac {3}{4},-\frac {3}{4},1;-\frac {1}{4},\frac {1}{4};(c+d x)^2\right )+3 \left (a+b \sin ^{-1}(c+d x)\right ) \left (5 \left (a+b \sin ^{-1}(c+d x)\right )+4 b (c+d x) \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};(c+d x)^2\right )\right )\right )}{105 d e^5 (c+d x)^4} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^(9/2),x]
[Out]
(-2*Sqrt[e*(c + d*x)]*(3*(a + b*ArcSin[c + d*x])*(5*(a + b*ArcSin[c + d*x]) + 4*b*(c + d*x)*Hypergeometric2F1[
-5/4, 1/2, -1/4, (c + d*x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{-3/4, -3/4, 1}, {-1/4, 1/4}, (c + d*x)^2
]))/(105*d*e^5*(c + d*x)^4)
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{5} e^{5} x^{5} + 5 \, c d^{4} e^{5} x^{4} + 10 \, c^{2} d^{3} e^{5} x^{3} + 10 \, c^{3} d^{2} e^{5} x^{2} + 5 \, c^{4} d e^{5} x + c^{5} e^{5}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(9/2),x, algorithm="fricas")
[Out]
integral((b^2*arcsin(d*x + c)^2 + 2*a*b*arcsin(d*x + c) + a^2)*sqrt(d*e*x + c*e)/(d^5*e^5*x^5 + 5*c*d^4*e^5*x^
4 + 10*c^2*d^3*e^5*x^3 + 10*c^3*d^2*e^5*x^2 + 5*c^4*d*e^5*x + c^5*e^5), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac {9}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(9/2),x, algorithm="giac")
[Out]
integrate((b*arcsin(d*x + c) + a)^2/(d*e*x + c*e)^(9/2), x)
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maple [F] time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {9}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(9/2),x)
[Out]
int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(9/2),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(9/2),x, algorithm="maxima")
[Out]
-1/210*(60*b^2*sqrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 - (d^4*e^5*x^3 + 3*c*d^3*e^5*x
^2 + 3*c^2*d^2*e^5*x + c^3*d*e^5)*(2940*a*b*d^2*sqrt(e)*integrate(1/7*sqrt(d*x + c)*x^2*arctan(d*x/(sqrt(d*x +
c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^7*e^5*x^7 + 7*c*d^6*e^5*x^6 + 21*c^
2*d^5*e^5*x^5 + 35*c^3*d^4*e^5*x^4 + 35*c^4*d^3*e^5*x^3 - d^5*e^5*x^5 + 21*c^5*d^2*e^5*x^2 - 5*c*d^4*e^5*x^4 +
7*c^6*d*e^5*x - 10*c^2*d^3*e^5*x^3 + c^7*e^5 - 10*c^3*d^2*e^5*x^2 - 5*c^4*d*e^5*x - c^5*e^5), x) + 5880*a*b*c
*d*sqrt(e)*integrate(1/7*sqrt(d*x + c)*x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c +
1)*sqrt(-d*x - c + 1)))/(d^7*e^5*x^7 + 7*c*d^6*e^5*x^6 + 21*c^2*d^5*e^5*x^5 + 35*c^3*d^4*e^5*x^4 + 35*c^4*d^3
*e^5*x^3 - d^5*e^5*x^5 + 21*c^5*d^2*e^5*x^2 - 5*c*d^4*e^5*x^4 + 7*c^6*d*e^5*x - 10*c^2*d^3*e^5*x^3 + c^7*e^5 -
10*c^3*d^2*e^5*x^2 - 5*c^4*d*e^5*x - c^5*e^5), x) + 2940*a*b*c^2*sqrt(e)*integrate(1/7*sqrt(d*x + c)*arctan(d
*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^7*e^5*x^7 + 7*c*d^6*e
^5*x^6 + 21*c^2*d^5*e^5*x^5 + 35*c^3*d^4*e^5*x^4 + 35*c^4*d^3*e^5*x^3 - d^5*e^5*x^5 + 21*c^5*d^2*e^5*x^2 - 5*c
*d^4*e^5*x^4 + 7*c^6*d*e^5*x - 10*c^2*d^3*e^5*x^3 + c^7*e^5 - 10*c^3*d^2*e^5*x^2 - 5*c^4*d*e^5*x - c^5*e^5), x
) - 5*a^2*c^2*sqrt(e)*(42*arctan(sqrt(d*x + c))/e^5 + 21*log(sqrt(d*x + c) + 1)/e^5 - 21*log(sqrt(d*x + c) - 1
)/e^5 - 4*(7*(d*x + c)^2 + 3)/((d*x + c)^(7/2)*e^5))/d - 840*b^2*d*sqrt(e)*integrate(1/7*sqrt(d*x + c + 1)*sqr
t(d*x + c)*sqrt(-d*x - c + 1)*x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(
-d*x - c + 1)))/(d^7*e^5*x^7 + 7*c*d^6*e^5*x^6 + 21*c^2*d^5*e^5*x^5 + 35*c^3*d^4*e^5*x^4 + 35*c^4*d^3*e^5*x^3
- d^5*e^5*x^5 + 21*c^5*d^2*e^5*x^2 - 5*c*d^4*e^5*x^4 + 7*c^6*d*e^5*x - 10*c^2*d^3*e^5*x^3 + c^7*e^5 - 10*c^3*d
^2*e^5*x^2 - 5*c^4*d*e^5*x - c^5*e^5), x) - 840*b^2*c*sqrt(e)*integrate(1/7*sqrt(d*x + c + 1)*sqrt(d*x + c)*sq
rt(-d*x - c + 1)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))
/(d^7*e^5*x^7 + 7*c*d^6*e^5*x^6 + 21*c^2*d^5*e^5*x^5 + 35*c^3*d^4*e^5*x^4 + 35*c^4*d^3*e^5*x^3 - d^5*e^5*x^5 +
21*c^5*d^2*e^5*x^2 - 5*c*d^4*e^5*x^4 + 7*c^6*d*e^5*x - 10*c^2*d^3*e^5*x^3 + c^7*e^5 - 10*c^3*d^2*e^5*x^2 - 5*
c^4*d*e^5*x - c^5*e^5), x) + 2*a^2*c*sqrt(e)*(210*(c + 1)*arctan(sqrt(d*x + c))/e^5 + 105*(c - 1)*log(sqrt(d*x
+ c) + 1)/e^5 - 105*(c - 1)*log(sqrt(d*x + c) - 1)/e^5 + 4*(105*(d*x + c)^3 - 35*(d*x + c)^2*c + 21*d*x + 6*c
)/((d*x + c)^(7/2)*e^5))/d - 2940*a*b*sqrt(e)*integrate(1/7*sqrt(d*x + c)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-
d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^7*e^5*x^7 + 7*c*d^6*e^5*x^6 + 21*c^2*d^5*e^5*x^5
+ 35*c^3*d^4*e^5*x^4 + 35*c^4*d^3*e^5*x^3 - d^5*e^5*x^5 + 21*c^5*d^2*e^5*x^2 - 5*c*d^4*e^5*x^4 + 7*c^6*d*e^5*x
- 10*c^2*d^3*e^5*x^3 + c^7*e^5 - 10*c^3*d^2*e^5*x^2 - 5*c^4*d*e^5*x - c^5*e^5), x) - a^2*sqrt(e)*(210*(c^2 +
2*c + 1)*arctan(sqrt(d*x + c))/e^5 + 105*(c^2 - 2*c + 1)*log(sqrt(d*x + c) + 1)/e^5 - 105*(c^2 - 2*c + 1)*log(
sqrt(d*x + c) - 1)/e^5 + 4*(210*(d*x + c)^3*c - 35*(c^2 + 1)*(d*x + c)^2 + 42*(d*x + c)*c - 15*c^2)/((d*x + c)
^(7/2)*e^5))/d + 5*a^2*sqrt(e)*(42*arctan(sqrt(d*x + c))/e^5 + 21*log(sqrt(d*x + c) + 1)/e^5 - 21*log(sqrt(d*x
+ c) - 1)/e^5 - 4*(7*(d*x + c)^2 + 3)/((d*x + c)^(7/2)*e^5))/d)*sqrt(d*x + c))/((d^4*e^5*x^3 + 3*c*d^3*e^5*x^
2 + 3*c^2*d^2*e^5*x + c^3*d*e^5)*sqrt(d*x + c))
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^{9/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^(9/2),x)
[Out]
int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^(9/2), 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))**2/(d*e*x+c*e)**(9/2),x)
[Out]
Timed out
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