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

Optimal. Leaf size=130 \[ -\frac {16 b^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};(c+d x)^2\right )}{15 d e^3 \sqrt {e (c+d x)}}-\frac {8 b \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}} \]

[Out]

-2/5*(a+b*arcsin(d*x+c))^2/d/e/(e*(d*x+c))^(5/2)-8/15*b*(a+b*arcsin(d*x+c))*hypergeom([-3/4, 1/2],[1/4],(d*x+c
)^2)/d/e^2/(e*(d*x+c))^(3/2)-16/15*b^2*HypergeometricPFQ([-1/4, -1/4, 1],[1/4, 3/4],(d*x+c)^2)/d/e^3/(e*(d*x+c
))^(1/2)

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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} \[ -\frac {16 b^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};(c+d x)^2\right )}{15 d e^3 \sqrt {e (c+d x)}}-\frac {8 b \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSin[c + d*x])^2)/(5*d*e*(e*(c + d*x))^(5/2)) - (8*b*(a + b*ArcSin[c + d*x])*Hypergeometric2F1[-3
/4, 1/2, 1/4, (c + d*x)^2])/(15*d*e^2*(e*(c + d*x))^(3/2)) - (16*b^2*HypergeometricPFQ[{-1/4, -1/4, 1}, {1/4,
3/4}, (c + d*x)^2])/(15*d*e^3*Sqrt[e*(c + d*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 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)^{7/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{(e x)^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{(e x)^{5/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}}-\frac {8 b \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};(c+d x)^2\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac {16 b^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};(c+d x)^2\right )}{15 d e^3 \sqrt {e (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 106, normalized size = 0.82 \[ -\frac {2 \left (8 b^2 (c+d x)^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};(c+d x)^2\right )+\left (a+b \sin ^{-1}(c+d x)\right ) \left (3 \left (a+b \sin ^{-1}(c+d x)\right )+4 b (c+d x) \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};(c+d x)^2\right )\right )\right )}{15 d e (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((a + b*ArcSin[c + d*x])*(3*(a + b*ArcSin[c + d*x]) + 4*b*(c + d*x)*Hypergeometric2F1[-3/4, 1/2, 1/4, (c +
d*x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{-1/4, -1/4, 1}, {1/4, 3/4}, (c + d*x)^2]))/(15*d*e*(e*(c + d*
x))^(5/2))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(7/2),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^4*e^4*x^4 + 4*c*d^3*e^4*x^
3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-1/30*(12*b^2*sqrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 - (d^3*e^4*x^2 + 2*c*d^2*e^4*x
+ c^2*d*e^4)*(300*a*b*d^2*sqrt(e)*integrate(1/5*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^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + 15*c^2*d^4*e^4*x^4 + 20*c^3
*d^3*e^4*x^3 + 15*c^4*d^2*e^4*x^2 - d^4*e^4*x^4 + 6*c^5*d*e^4*x - 4*c*d^3*e^4*x^3 + c^6*e^4 - 6*c^2*d^2*e^4*x^
2 - 4*c^3*d*e^4*x - c^4*e^4), x) + 600*a*b*c*d*sqrt(e)*integrate(1/5*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^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + 15*c^2*d
^4*e^4*x^4 + 20*c^3*d^3*e^4*x^3 + 15*c^4*d^2*e^4*x^2 - d^4*e^4*x^4 + 6*c^5*d*e^4*x - 4*c*d^3*e^4*x^3 + c^6*e^4
- 6*c^2*d^2*e^4*x^2 - 4*c^3*d*e^4*x - c^4*e^4), x) + 300*a*b*c^2*sqrt(e)*integrate(1/5*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^6*e^4*x^6 + 6*c*d^5*e
^4*x^5 + 15*c^2*d^4*e^4*x^4 + 20*c^3*d^3*e^4*x^3 + 15*c^4*d^2*e^4*x^2 - d^4*e^4*x^4 + 6*c^5*d*e^4*x - 4*c*d^3*
e^4*x^3 + c^6*e^4 - 6*c^2*d^2*e^4*x^2 - 4*c^3*d*e^4*x - c^4*e^4), x) + 3*a^2*c^2*sqrt(e)*(10*arctan(sqrt(d*x +
c))/e^4 - 5*log(sqrt(d*x + c) + 1)/e^4 + 5*log(sqrt(d*x + c) - 1)/e^4 + 4*(5*(d*x + c)^2 + 1)/((d*x + c)^(5/2
)*e^4))/d - 120*b^2*d*sqrt(e)*integrate(1/5*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*x*arctan(d*x/(s
qrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^6*e^4*x^6 + 6*c*d^5*e^4*x^
5 + 15*c^2*d^4*e^4*x^4 + 20*c^3*d^3*e^4*x^3 + 15*c^4*d^2*e^4*x^2 - d^4*e^4*x^4 + 6*c^5*d*e^4*x - 4*c*d^3*e^4*x
^3 + c^6*e^4 - 6*c^2*d^2*e^4*x^2 - 4*c^3*d*e^4*x - c^4*e^4), x) - 120*b^2*c*sqrt(e)*integrate(1/5*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^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + 15*c^2*d^4*e^4*x^4 + 20*c^3*d^3*e^4*x^3 + 15*c^4*d^2*e
^4*x^2 - d^4*e^4*x^4 + 6*c^5*d*e^4*x - 4*c*d^3*e^4*x^3 + c^6*e^4 - 6*c^2*d^2*e^4*x^2 - 4*c^3*d*e^4*x - c^4*e^4
), x) - 2*a^2*c*sqrt(e)*(30*(c + 1)*arctan(sqrt(d*x + c))/e^4 - 15*(c - 1)*log(sqrt(d*x + c) + 1)/e^4 + 15*(c
- 1)*log(sqrt(d*x + c) - 1)/e^4 + 4*(15*(d*x + c)^2*c - 5*d*x - 2*c)/((d*x + c)^(5/2)*e^4))/d - 300*a*b*sqrt(e
)*integrate(1/5*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^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + 15*c^2*d^4*e^4*x^4 + 20*c^3*d^3*e^4*x^3 + 15*c^4*d^2*e^4*x^2 -
d^4*e^4*x^4 + 6*c^5*d*e^4*x - 4*c*d^3*e^4*x^3 + c^6*e^4 - 6*c^2*d^2*e^4*x^2 - 4*c^3*d*e^4*x - c^4*e^4), x) + a
^2*sqrt(e)*(30*(c^2 + 2*c + 1)*arctan(sqrt(d*x + c))/e^4 - 15*(c^2 - 2*c + 1)*log(sqrt(d*x + c) + 1)/e^4 + 15*
(c^2 - 2*c + 1)*log(sqrt(d*x + c) - 1)/e^4 + 4*(15*(c^2 + 1)*(d*x + c)^2 - 10*(d*x + c)*c + 3*c^2)/((d*x + c)^
(5/2)*e^4))/d - 3*a^2*sqrt(e)*(10*arctan(sqrt(d*x + c))/e^4 - 5*log(sqrt(d*x + c) + 1)/e^4 + 5*log(sqrt(d*x +
c) - 1)/e^4 + 4*(5*(d*x + c)^2 + 1)/((d*x + c)^(5/2)*e^4))/d)*sqrt(d*x + c))/((d^3*e^4*x^2 + 2*c*d^2*e^4*x + c
^2*d*e^4)*sqrt(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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