3.3.92 \(\int (c e+d e x)^{5/2} (a+b \text {ArcSin}(c+d x))^2 \, dx\) [292]
Optimal. Leaf size=130 \[ \frac {2 (e (c+d x))^{7/2} (a+b \text {ArcSin}(c+d x))^2}{7 d e}-\frac {8 b (e (c+d x))^{9/2} (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right )}{63 d e^2}+\frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )}{693 d e^3} \]
[Out]
2/7*(e*(d*x+c))^(7/2)*(a+b*arcsin(d*x+c))^2/d/e-8/63*b*(e*(d*x+c))^(9/2)*(a+b*arcsin(d*x+c))*hypergeom([1/2, 9
/4],[13/4],(d*x+c)^2)/d/e^2+16/693*b^2*(e*(d*x+c))^(11/2)*hypergeom([1, 11/4, 11/4],[13/4, 15/4],(d*x+c)^2)/d/
e^3
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Rubi [A]
time = 0.14, 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 = {4889, 4723,
4805} \begin {gather*} \frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )}{693 d e^3}-\frac {8 b (e (c+d x))^{9/2} \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right ) (a+b \text {ArcSin}(c+d x))}{63 d e^2}+\frac {2 (e (c+d x))^{7/2} (a+b \text {ArcSin}(c+d x))^2}{7 d e} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(c*e + d*e*x)^(5/2)*(a + b*ArcSin[c + d*x])^2,x]
[Out]
(2*(e*(c + d*x))^(7/2)*(a + b*ArcSin[c + d*x])^2)/(7*d*e) - (8*b*(e*(c + d*x))^(9/2)*(a + b*ArcSin[c + d*x])*H
ypergeometric2F1[1/2, 9/4, 13/4, (c + d*x)^2])/(63*d*e^2) + (16*b^2*(e*(c + d*x))^(11/2)*HypergeometricPFQ[{1,
11/4, 11/4}, {13/4, 15/4}, (c + d*x)^2])/(693*d*e^3)
Rule 4723
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_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, c^2*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d +
e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e,
f, m}, x] && EqQ[c^2*d + e, 0] && !IntegerQ[m]
Rule 4889
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)^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int (e x)^{5/2} \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{7/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{7 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{7/2} \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{7 d e}\\ &=\frac {2 (e (c+d x))^{7/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{7 d e}-\frac {8 b (e (c+d x))^{9/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right )}{63 d e^2}+\frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )}{693 d e^3}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 106, normalized size = 0.82 \begin {gather*} \frac {2 (e (c+d x))^{7/2} \left (99 (a+b \text {ArcSin}(c+d x))^2-44 b (c+d x) (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right )+8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )\right )}{693 d e} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(c*e + d*e*x)^(5/2)*(a + b*ArcSin[c + d*x])^2,x]
[Out]
(2*(e*(c + d*x))^(7/2)*(99*(a + b*ArcSin[c + d*x])^2 - 44*b*(c + d*x)*(a + b*ArcSin[c + d*x])*Hypergeometric2F
1[1/2, 9/4, 13/4, (c + d*x)^2] + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, (c + d*x)^
2]))/(693*d*e)
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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{\frac {5}{2}} \left (a +b \arcsin \left (d x +c \right )\right )^{2}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*e*x+c*e)^(5/2)*(a+b*arcsin(d*x+c))^2,x)
[Out]
int((d*e*x+c*e)^(5/2)*(a+b*arcsin(d*x+c))^2,x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)^(5/2)*(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")
[Out]
1/210*(60*(b^2*d^3*x^3*e^2 + 3*b^2*c*d^2*x^2*e^2 + 3*b^2*c^2*d*x*e^2 + b^2*c^3*e^2)*sqrt(d*x + c)*arctan2(d*x
+ c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2*e^(1/2) + (2940*a*b*d^4*e^(5/2)*integrate(1/7*sqrt(d*x + c)*x^4*a
rctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*
x + c^2 - 1), x) + 11760*a*b*c*d^3*e^(5/2)*integrate(1/7*sqrt(d*x + c)*x^3*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^2*x^2 + 2*c*d*x + c^2 - 1), x) + 17640*a*b*c^2*d
^2*e^(5/2)*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^2*x^2 + 2*c*d*x + c^2 - 1), x) + 11760*a*b*c^3*d*e^(5/2)*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^2*x^2
+ 2*c*d*x + c^2 - 1), x) + 2940*a*b*c^4*e^(5/2)*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^2*x^2 + 2*c*d*x + c^2 - 1), x) + 105*a^2*c^4*(2
*arctan(sqrt(d*x + c)) - log(sqrt(d*x + c) + 1) + log(sqrt(d*x + c) - 1))*e^(5/2)/d + 840*b^2*d^3*e^(5/2)*inte
grate(1/7*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*x^3*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^2*x^2 + 2*c*d*x + c^2 - 1), x) + 2520*b^2*c*d^2*e^(5/2)*in
tegrate(1/7*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*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^2*x^2 + 2*c*d*x + c^2 - 1), x) + 2520*b^2*c^2*d*e^(5/2)*
integrate(1/7*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)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 840*b^2*c^3*e^(5/2)*int
egrate(1/7*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^2*x^2 + 2*c*d*x + c^2 - 1), x) - 420*(2*(c + 1)*arctan(sqrt(d
*x + c)) - (c - 1)*log(sqrt(d*x + c) + 1) + (c - 1)*log(sqrt(d*x + c) - 1) - 4*sqrt(d*x + c))*a^2*c^3*e^(5/2)/
d - 2940*a*b*d^2*e^(5/2)*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^2*x^2 + 2*c*d*x + c^2 - 1), x) - 5880*a*b*c*d*e^(5/2)*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^2*x^2 + 2*c*d*x + c^2 - 1), x) - 2940*a*b*c^2*e^(5/2)*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^2*x^2 + 2*c*d*x + c^2 - 1), x) + 210
*(6*(c^2 + 2*c + 1)*arctan(sqrt(d*x + c)) - 3*(c^2 - 2*c + 1)*log(sqrt(d*x + c) + 1) + 3*(c^2 - 2*c + 1)*log(s
qrt(d*x + c) - 1) + 4*(d*x + c)^(3/2) - 24*sqrt(d*x + c)*c)*a^2*c^2*e^(5/2)/d - 105*a^2*c^2*(2*arctan(sqrt(d*x
+ c)) - log(sqrt(d*x + c) + 1) + log(sqrt(d*x + c) - 1))*e^(5/2)/d + 84*(4*(d*x + c)^(5/2) - 20*(d*x + c)^(3/
2)*c - 10*(c^3 + 3*c^2 + 3*c + 1)*arctan(sqrt(d*x + c)) + 5*(c^3 - 3*c^2 + 3*c - 1)*log(sqrt(d*x + c) + 1) - 5
*(c^3 - 3*c^2 + 3*c - 1)*log(sqrt(d*x + c) - 1) + 20*(3*c^2 + 1)*sqrt(d*x + c))*a^2*c*e^(5/2)/d + 210*(2*(c +
1)*arctan(sqrt(d*x + c)) - (c - 1)*log(sqrt(d*x + c) + 1) + (c - 1)*log(sqrt(d*x + c) - 1) - 4*sqrt(d*x + c))*
a^2*c*e^(5/2)/d + (60*(d*x + c)^(7/2) - 336*(d*x + c)^(5/2)*c + 140*(6*c^2 + 1)*(d*x + c)^(3/2) + 210*(c^4 + 4
*c^3 + 6*c^2 + 4*c + 1)*arctan(sqrt(d*x + c)) - 105*(c^4 - 4*c^3 + 6*c^2 - 4*c + 1)*log(sqrt(d*x + c) + 1) + 1
05*(c^4 - 4*c^3 + 6*c^2 - 4*c + 1)*log(sqrt(d*x + c) - 1) - 1680*(c^3 + c)*sqrt(d*x + c))*a^2*e^(5/2)/d - 35*(
6*(c^2 + 2*c + 1)*arctan(sqrt(d*x + c)) - 3*(c^2 - 2*c + 1)*log(sqrt(d*x + c) + 1) + 3*(c^2 - 2*c + 1)*log(sqr
t(d*x + c) - 1) + 4*(d*x + c)^(3/2) - 24*sqrt(d*x + c)*c)*a^2*e^(5/2)/d)*d)/d
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)^(5/2)*(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")
[Out]
integral(((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*arcsin(d*x + c)^2*e^2 + 2*(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2
)*arcsin(d*x + c)*e^2 + (a^2*d^2*x^2 + 2*a^2*c*d*x + a^2*c^2)*e^2)*sqrt(d*x + c)*e^(1/2), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \left (c + d x\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)**(5/2)*(a+b*asin(d*x+c))**2,x)
[Out]
Integral((e*(c + d*x))**(5/2)*(a + b*asin(c + d*x))**2, x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)^(5/2)*(a+b*arcsin(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((d*e*x + c*e)^(5/2)*(b*arcsin(d*x + c) + a)^2, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^{5/2}\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*e + d*e*x)^(5/2)*(a + b*asin(c + d*x))^2,x)
[Out]
int((c*e + d*e*x)^(5/2)*(a + b*asin(c + d*x))^2, x)
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