3.843 \(\int \frac{(d+e x)^{3/2} (15 d^2+20 d e x+8 e^2 x^2)}{\sqrt{a+b x}} \, dx\)

Optimal. Leaf size=240 \[ \frac{\sqrt{a+b x} (d+e x)^{3/2} \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right )}{12 b^3}+\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right )}{8 b^4}+\frac{(b d-a e)^2 \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{9/2} \sqrt{e}}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\sqrt{a+b x} (d+e x)^{5/2} (17 b d-13 a e)}{3 b^2} \]

[Out]

((b*d - a*e)*(73*b^2*d^2 - 90*a*b*d*e + 35*a^2*e^2)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^4) + ((73*b^2*d^2 - 90*a
*b*d*e + 35*a^2*e^2)*Sqrt[a + b*x]*(d + e*x)^(3/2))/(12*b^3) + ((17*b*d - 13*a*e)*Sqrt[a + b*x]*(d + e*x)^(5/2
))/(3*b^2) + (2*e*(a + b*x)^(3/2)*(d + e*x)^(5/2))/b^2 + ((b*d - a*e)^2*(73*b^2*d^2 - 90*a*b*d*e + 35*a^2*e^2)
*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(9/2)*Sqrt[e])

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Rubi [A]  time = 0.225081, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {951, 80, 50, 63, 217, 206} \[ \frac{\sqrt{a+b x} (d+e x)^{3/2} \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right )}{12 b^3}+\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right )}{8 b^4}+\frac{(b d-a e)^2 \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{9/2} \sqrt{e}}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\sqrt{a+b x} (d+e x)^{5/2} (17 b d-13 a e)}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^(3/2)*(15*d^2 + 20*d*e*x + 8*e^2*x^2))/Sqrt[a + b*x],x]

[Out]

((b*d - a*e)*(73*b^2*d^2 - 90*a*b*d*e + 35*a^2*e^2)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^4) + ((73*b^2*d^2 - 90*a
*b*d*e + 35*a^2*e^2)*Sqrt[a + b*x]*(d + e*x)^(3/2))/(12*b^3) + ((17*b*d - 13*a*e)*Sqrt[a + b*x]*(d + e*x)^(5/2
))/(3*b^2) + (2*e*(a + b*x)^(3/2)*(d + e*x)^(5/2))/b^2 + ((b*d - a*e)^2*(73*b^2*d^2 - 90*a*b*d*e + 35*a^2*e^2)
*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(9/2)*Sqrt[e])

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^{3/2} \left (15 d^2+20 d e x+8 e^2 x^2\right )}{\sqrt{a+b x}} \, dx &=\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\int \frac{(d+e x)^{3/2} \left (4 e \left (15 b^2 d^2-3 a b d e-5 a^2 e^2\right )+4 b e^2 (17 b d-13 a e) x\right )}{\sqrt{a+b x}} \, dx}{4 b^2 e}\\ &=\frac{(17 b d-13 a e) \sqrt{a+b x} (d+e x)^{5/2}}{3 b^2}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \int \frac{(d+e x)^{3/2}}{\sqrt{a+b x}} \, dx}{6 b^2}\\ &=\frac{\left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^3}+\frac{(17 b d-13 a e) \sqrt{a+b x} (d+e x)^{5/2}}{3 b^2}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\left ((b d-a e) \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x}} \, dx}{8 b^3}\\ &=\frac{(b d-a e) \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{8 b^4}+\frac{\left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^3}+\frac{(17 b d-13 a e) \sqrt{a+b x} (d+e x)^{5/2}}{3 b^2}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\left ((b d-a e)^2 \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{16 b^4}\\ &=\frac{(b d-a e) \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{8 b^4}+\frac{\left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^3}+\frac{(17 b d-13 a e) \sqrt{a+b x} (d+e x)^{5/2}}{3 b^2}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\left ((b d-a e)^2 \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{8 b^5}\\ &=\frac{(b d-a e) \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{8 b^4}+\frac{\left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^3}+\frac{(17 b d-13 a e) \sqrt{a+b x} (d+e x)^{5/2}}{3 b^2}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{\left ((b d-a e)^2 \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{8 b^5}\\ &=\frac{(b d-a e) \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{8 b^4}+\frac{\left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^3}+\frac{(17 b d-13 a e) \sqrt{a+b x} (d+e x)^{5/2}}{3 b^2}+\frac{2 e (a+b x)^{3/2} (d+e x)^{5/2}}{b^2}+\frac{(b d-a e)^2 \left (73 b^2 d^2-90 a b d e+35 a^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{9/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 0.909865, size = 204, normalized size = 0.85 \[ \frac{\sqrt{d+e x} \left (\sqrt{a+b x} \left (5 a^2 b e^2 (89 d+14 e x)-105 a^3 e^3-a b^2 e \left (725 d^2+292 d e x+56 e^2 x^2\right )+b^3 \left (466 d^2 e x+501 d^3+232 d e^2 x^2+48 e^3 x^3\right )\right )+\frac{3 \left (35 a^2 e^2-90 a b d e+73 b^2 d^2\right ) (b d-a e)^{3/2} \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{e} \sqrt{\frac{b (d+e x)}{b d-a e}}}\right )}{24 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^(3/2)*(15*d^2 + 20*d*e*x + 8*e^2*x^2))/Sqrt[a + b*x],x]

[Out]

(Sqrt[d + e*x]*(Sqrt[a + b*x]*(-105*a^3*e^3 + 5*a^2*b*e^2*(89*d + 14*e*x) - a*b^2*e*(725*d^2 + 292*d*e*x + 56*
e^2*x^2) + b^3*(501*d^3 + 466*d^2*e*x + 232*d*e^2*x^2 + 48*e^3*x^3)) + (3*(b*d - a*e)^(3/2)*(73*b^2*d^2 - 90*a
*b*d*e + 35*a^2*e^2)*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/(Sqrt[e]*Sqrt[(b*(d + e*x))/(b*d - a*e)
])))/(24*b^4)

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Maple [B]  time = 0.375, size = 571, normalized size = 2.4 \begin{align*}{\frac{1}{48\,{b}^{4}}\sqrt{ex+d}\sqrt{bx+a} \left ( 96\,{x}^{3}{b}^{3}{e}^{3}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}-112\,{x}^{2}a{b}^{2}{e}^{3}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+464\,{x}^{2}{b}^{3}d{e}^{2}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+105\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{4}{e}^{4}-480\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{3}bd{e}^{3}+864\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}{b}^{2}{d}^{2}{e}^{2}-708\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) a{b}^{3}{d}^{3}e+219\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{4}{d}^{4}+140\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}x{a}^{2}b{e}^{3}-584\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}xa{b}^{2}d{e}^{2}+932\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}x{b}^{3}{d}^{2}e-210\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{a}^{3}{e}^{3}+890\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{a}^{2}bd{e}^{2}-1450\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }a{b}^{2}{d}^{2}e+1002\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{b}^{3}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x)

[Out]

1/48*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(96*x^3*b^3*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-112*x^2*a*b^2*e^3*((b*x+a
)*(e*x+d))^(1/2)*(b*e)^(1/2)+464*x^2*b^3*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+105*ln(1/2*(2*b*x*e+2*((b*x
+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^4*e^4-480*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*b*d*e^3+864*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/
(b*e)^(1/2))*a^2*b^2*d^2*e^2-708*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a
*b^3*d^3*e+219*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^4*d^4+140*((b*x+a
)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*a^2*b*e^3-584*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*a*b^2*d*e^2+932*((b*x+a)*(e
*x+d))^(1/2)*(b*e)^(1/2)*x*b^3*d^2*e-210*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a^3*e^3+890*(b*e)^(1/2)*((b*x+a)*
(e*x+d))^(1/2)*a^2*b*d*e^2-1450*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b^2*d^2*e+1002*(b*e)^(1/2)*((b*x+a)*(e*x
+d))^(1/2)*b^3*d^3)/b^4/((b*x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12577, size = 1250, normalized size = 5.21 \begin{align*} \left [\frac{3 \,{\left (73 \, b^{4} d^{4} - 236 \, a b^{3} d^{3} e + 288 \, a^{2} b^{2} d^{2} e^{2} - 160 \, a^{3} b d e^{3} + 35 \, a^{4} e^{4}\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \,{\left (48 \, b^{4} e^{4} x^{3} + 501 \, b^{4} d^{3} e - 725 \, a b^{3} d^{2} e^{2} + 445 \, a^{2} b^{2} d e^{3} - 105 \, a^{3} b e^{4} + 8 \,{\left (29 \, b^{4} d e^{3} - 7 \, a b^{3} e^{4}\right )} x^{2} + 2 \,{\left (233 \, b^{4} d^{2} e^{2} - 146 \, a b^{3} d e^{3} + 35 \, a^{2} b^{2} e^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{96 \, b^{5} e}, -\frac{3 \,{\left (73 \, b^{4} d^{4} - 236 \, a b^{3} d^{3} e + 288 \, a^{2} b^{2} d^{2} e^{2} - 160 \, a^{3} b d e^{3} + 35 \, a^{4} e^{4}\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \,{\left (48 \, b^{4} e^{4} x^{3} + 501 \, b^{4} d^{3} e - 725 \, a b^{3} d^{2} e^{2} + 445 \, a^{2} b^{2} d e^{3} - 105 \, a^{3} b e^{4} + 8 \,{\left (29 \, b^{4} d e^{3} - 7 \, a b^{3} e^{4}\right )} x^{2} + 2 \,{\left (233 \, b^{4} d^{2} e^{2} - 146 \, a b^{3} d e^{3} + 35 \, a^{2} b^{2} e^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{48 \, b^{5} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(73*b^4*d^4 - 236*a*b^3*d^3*e + 288*a^2*b^2*d^2*e^2 - 160*a^3*b*d*e^3 + 35*a^4*e^4)*sqrt(b*e)*log(8*b
^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8
*(b^2*d*e + a*b*e^2)*x) + 4*(48*b^4*e^4*x^3 + 501*b^4*d^3*e - 725*a*b^3*d^2*e^2 + 445*a^2*b^2*d*e^3 - 105*a^3*
b*e^4 + 8*(29*b^4*d*e^3 - 7*a*b^3*e^4)*x^2 + 2*(233*b^4*d^2*e^2 - 146*a*b^3*d*e^3 + 35*a^2*b^2*e^4)*x)*sqrt(b*
x + a)*sqrt(e*x + d))/(b^5*e), -1/48*(3*(73*b^4*d^4 - 236*a*b^3*d^3*e + 288*a^2*b^2*d^2*e^2 - 160*a^3*b*d*e^3
+ 35*a^4*e^4)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2
+ a*b*d*e + (b^2*d*e + a*b*e^2)*x)) - 2*(48*b^4*e^4*x^3 + 501*b^4*d^3*e - 725*a*b^3*d^2*e^2 + 445*a^2*b^2*d*e^
3 - 105*a^3*b*e^4 + 8*(29*b^4*d*e^3 - 7*a*b^3*e^4)*x^2 + 2*(233*b^4*d^2*e^2 - 146*a*b^3*d*e^3 + 35*a^2*b^2*e^4
)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^5*e)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(8*e**2*x**2+20*d*e*x+15*d**2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.46333, size = 975, normalized size = 4.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/48*(720*((b^2*d - a*b*e)*e^(-1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b
*e)))/sqrt(b) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a))*d^3*abs(b)/b^2 - 56*(sqrt(b^2*d + (b*x + a)
*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)*e^(-4)/b^7) - 3*(b^7*d^
2*e^2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)*e^(-4)/b^7) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*e^
(-5/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(3/2))*d*abs(b)*e^2/b^
2 - 35*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e^(-2)/b^4 + (b*d*e - 5*a*e^2)*e^(-4)/b
^4) + (b^2*d^2 + 2*a*b*d*e - 3*a^2*e^2)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x +
a)*b*e - a*b*e)))/b^(7/2))*d^2*abs(b)*e/b^3 - 2*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)
*(6*(b*x + a)/b^3 + (b^12*d*e^5 - 25*a*b^11*e^6)*e^(-6)/b^14) - (5*b^13*d^2*e^4 + 14*a*b^12*d*e^5 - 163*a^2*b^
11*e^6)*e^(-6)/b^14) + 3*(5*b^14*d^3*e^3 + 9*a*b^13*d^2*e^4 + 15*a^2*b^12*d*e^5 - 93*a^3*b^11*e^6)*e^(-6)/b^14
)*sqrt(b*x + a) + 3*(5*b^4*d^4 + 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 20*a^3*b*d*e^3 - 35*a^4*e^4)*e^(-7/2)*log
(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(5/2))*abs(b)*e^3/b^2)/b