3.297 \(\int \frac {(a+b \sin ^{-1}(c+d x))^2}{(c e+d e x)^{5/2}} \, dx\)
Optimal. Leaf size=130 \[ \frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}-\frac {8 b \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}} \]
[Out]
-2/3*(a+b*arcsin(d*x+c))^2/d/e/(e*(d*x+c))^(3/2)-8/3*b*(a+b*arcsin(d*x+c))*hypergeom([-1/4, 1/2],[3/4],(d*x+c)
^2)/d/e^2/(e*(d*x+c))^(1/2)+16/3*b^2*HypergeometricPFQ([1/4, 1/4, 1],[3/4, 5/4],(d*x+c)^2)*(e*(d*x+c))^(1/2)/d
/e^3
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Rubi [A] time = 0.21, 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} \[ \frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}-\frac {8 b \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^(5/2),x]
[Out]
(-2*(a + b*ArcSin[c + d*x])^2)/(3*d*e*(e*(c + d*x))^(3/2)) - (8*b*(a + b*ArcSin[c + d*x])*Hypergeometric2F1[-1
/4, 1/2, 3/4, (c + d*x)^2])/(3*d*e^2*Sqrt[e*(c + d*x)]) + (16*b^2*Sqrt[e*(c + d*x)]*HypergeometricPFQ[{1/4, 1/
4, 1}, {3/4, 5/4}, (c + d*x)^2])/(3*d*e^3)
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)^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{(e x)^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{(e x)^{3/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}}-\frac {8 b \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right )}{3 d e^2 \sqrt {e (c+d x)}}+\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 102, normalized size = 0.78 \[ -\frac {2 \left (4 b (c+d x) \left (\, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )-2 b (c+d x) \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )\right )+\left (a+b \sin ^{-1}(c+d x)\right )^2\right )}{3 d e (e (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^(5/2),x]
[Out]
(-2*((a + b*ArcSin[c + d*x])^2 + 4*b*(c + d*x)*((a + b*ArcSin[c + d*x])*Hypergeometric2F1[-1/4, 1/2, 3/4, (c +
d*x)^2] - 2*b*(c + d*x)*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, (c + d*x)^2])))/(3*d*e*(e*(c + d*x))^(3/
2))
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fricas [F] time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),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^3*e^3*x^3 + 3*c*d^2*e^3*x^
2 + 3*c^2*d*e^3*x + c^3*e^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),x, algorithm="giac")
[Out]
integrate((b*arcsin(d*x + c) + a)^2/(d*e*x + c*e)^(5/2), x)
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maple [F] time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),x)
[Out]
int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),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)^(5/2),x, algorithm="maxima")
[Out]
-1/6*(4*b^2*sqrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 - (d^2*e^3*x + c*d*e^3)*(36*a*b*d
^2*sqrt(e)*integrate(1/3*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^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + 10*c^2*d^3*e^3*x^3 + 10*c^3*d^2*e^3*x^2 + 5*c^4*d*
e^3*x - d^3*e^3*x^3 + c^5*e^3 - 3*c*d^2*e^3*x^2 - 3*c^2*d*e^3*x - c^3*e^3), x) + 72*a*b*c*d*sqrt(e)*integrate(
1/3*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^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + 10*c^2*d^3*e^3*x^3 + 10*c^3*d^2*e^3*x^2 + 5*c^4*d*e^3*x - d^3*e^3*x^3 + c
^5*e^3 - 3*c*d^2*e^3*x^2 - 3*c^2*d*e^3*x - c^3*e^3), x) + 36*a*b*c^2*sqrt(e)*integrate(1/3*sqrt(d*x + c)*arcta
n(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^5*e^3*x^5 + 5*c*d^
4*e^3*x^4 + 10*c^2*d^3*e^3*x^3 + 10*c^3*d^2*e^3*x^2 + 5*c^4*d*e^3*x - d^3*e^3*x^3 + c^5*e^3 - 3*c*d^2*e^3*x^2
- 3*c^2*d*e^3*x - c^3*e^3), x) - a^2*c^2*sqrt(e)*(6*arctan(sqrt(d*x + c))/e^3 + 3*log(sqrt(d*x + c) + 1)/e^3 -
3*log(sqrt(d*x + c) - 1)/e^3 - 4/((d*x + c)^(3/2)*e^3))/d - 24*b^2*d*sqrt(e)*integrate(1/3*sqrt(d*x + c + 1)*
sqrt(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)*sq
rt(-d*x - c + 1)))/(d^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + 10*c^2*d^3*e^3*x^3 + 10*c^3*d^2*e^3*x^2 + 5*c^4*d*e^3*x -
d^3*e^3*x^3 + c^5*e^3 - 3*c*d^2*e^3*x^2 - 3*c^2*d*e^3*x - c^3*e^3), x) - 24*b^2*c*sqrt(e)*integrate(1/3*sqrt(d
*x + c + 1)*sqrt(d*x + c)*sqrt(-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^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + 10*c^2*d^3*e^3*x^3 + 10*c^3*d^2*e^3*x^2 + 5*c^4*
d*e^3*x - d^3*e^3*x^3 + c^5*e^3 - 3*c*d^2*e^3*x^2 - 3*c^2*d*e^3*x - c^3*e^3), x) + 2*a^2*c*sqrt(e)*(6*(c + 1)*
arctan(sqrt(d*x + c))/e^3 + 3*(c - 1)*log(sqrt(d*x + c) + 1)/e^3 - 3*(c - 1)*log(sqrt(d*x + c) - 1)/e^3 + 4*(3
*d*x + 2*c)/((d*x + c)^(3/2)*e^3))/d - 36*a*b*sqrt(e)*integrate(1/3*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^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + 10*c^2*d^3*
e^3*x^3 + 10*c^3*d^2*e^3*x^2 + 5*c^4*d*e^3*x - d^3*e^3*x^3 + c^5*e^3 - 3*c*d^2*e^3*x^2 - 3*c^2*d*e^3*x - c^3*e
^3), x) - a^2*sqrt(e)*(6*(c^2 + 2*c + 1)*arctan(sqrt(d*x + c))/e^3 + 3*(c^2 - 2*c + 1)*log(sqrt(d*x + c) + 1)/
e^3 - 3*(c^2 - 2*c + 1)*log(sqrt(d*x + c) - 1)/e^3 + 4*(6*(d*x + c)*c - c^2)/((d*x + c)^(3/2)*e^3))/d + a^2*sq
rt(e)*(6*arctan(sqrt(d*x + c))/e^3 + 3*log(sqrt(d*x + c) + 1)/e^3 - 3*log(sqrt(d*x + c) - 1)/e^3 - 4/((d*x + c
)^(3/2)*e^3))/d)*sqrt(d*x + c))/((d^2*e^3*x + c*d*e^3)*sqrt(d*x + c))
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^(5/2),x)
[Out]
int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^(5/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}}{\left (e \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*asin(d*x+c))**2/(d*e*x+c*e)**(5/2),x)
[Out]
Integral((a + b*asin(c + d*x))**2/(e*(c + d*x))**(5/2), x)
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