3.591 \(\int \frac{x^3 \sqrt{a+b x}}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=218 \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (3 a^2 d^2-2 b d x (35 b c-31 a d)-100 a b c d+105 b^2 c^2\right )}{12 b d^4 (b c-a d)}+\frac{\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} d^{9/2}}-\frac{2 x^2 \sqrt{a+b x} (7 b c-6 a d)}{3 d^2 \sqrt{c+d x} (b c-a d)}-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}} \]

[Out]

(-2*x^3*Sqrt[a + b*x])/(3*d*(c + d*x)^(3/2)) - (2*(7*b*c - 6*a*d)*x^2*Sqrt[a + b*x])/(3*d^2*(b*c - a*d)*Sqrt[c
+ d*x]) - (Sqrt[a + b*x]*Sqrt[c + d*x]*(105*b^2*c^2 - 100*a*b*c*d + 3*a^2*d^2 - 2*b*d*(35*b*c - 31*a*d)*x))/(
12*b*d^4*(b*c - a*d)) + ((35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(4*b^(3/2)*d^(9/2))

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Rubi [A]  time = 0.187474, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {97, 150, 147, 63, 217, 206} \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (3 a^2 d^2-2 b d x (35 b c-31 a d)-100 a b c d+105 b^2 c^2\right )}{12 b d^4 (b c-a d)}+\frac{\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} d^{9/2}}-\frac{2 x^2 \sqrt{a+b x} (7 b c-6 a d)}{3 d^2 \sqrt{c+d x} (b c-a d)}-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sqrt[a + b*x])/(c + d*x)^(5/2),x]

[Out]

(-2*x^3*Sqrt[a + b*x])/(3*d*(c + d*x)^(3/2)) - (2*(7*b*c - 6*a*d)*x^2*Sqrt[a + b*x])/(3*d^2*(b*c - a*d)*Sqrt[c
+ d*x]) - (Sqrt[a + b*x]*Sqrt[c + d*x]*(105*b^2*c^2 - 100*a*b*c*d + 3*a^2*d^2 - 2*b*d*(35*b*c - 31*a*d)*x))/(
12*b*d^4*(b*c - a*d)) + ((35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(4*b^(3/2)*d^(9/2))

Rule 97

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

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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{x^3 \sqrt{a+b x}}{(c+d x)^{5/2}} \, dx &=-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}}+\frac{2 \int \frac{x^2 \left (3 a+\frac{7 b x}{2}\right )}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 d}\\ &=-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}}-\frac{2 (7 b c-6 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d) \sqrt{c+d x}}-\frac{4 \int \frac{x \left (-a (7 b c-6 a d)-\frac{1}{4} b (35 b c-31 a d) x\right )}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 d^2 (b c-a d)}\\ &=-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}}-\frac{2 (7 b c-6 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^2 c^2-100 a b c d+3 a^2 d^2-2 b d (35 b c-31 a d) x\right )}{12 b d^4 (b c-a d)}+\frac{\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b d^4}\\ &=-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}}-\frac{2 (7 b c-6 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^2 c^2-100 a b c d+3 a^2 d^2-2 b d (35 b c-31 a d) x\right )}{12 b d^4 (b c-a d)}+\frac{\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{4 b^2 d^4}\\ &=-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}}-\frac{2 (7 b c-6 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^2 c^2-100 a b c d+3 a^2 d^2-2 b d (35 b c-31 a d) x\right )}{12 b d^4 (b c-a d)}+\frac{\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 b^2 d^4}\\ &=-\frac{2 x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2}}-\frac{2 (7 b c-6 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^2 c^2-100 a b c d+3 a^2 d^2-2 b d (35 b c-31 a d) x\right )}{12 b d^4 (b c-a d)}+\frac{\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 1.14344, size = 249, normalized size = 1.14 \[ \frac{\frac{3 (c+d x)^2 \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \left (\sqrt{a+b x} (b c-a d) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )-\sqrt{d} (a+b x) \sqrt{b c-a d} \sqrt{\frac{b (c+d x)}{b c-a d}}\right )}{d^{7/2} (b c-a d)^{3/2} \sqrt{\frac{b (c+d x)}{b c-a d}}}+\frac{2 c (a+b x)^2 (3 a d (c+2 d x)-7 b c (5 c+6 d x))}{d^2 (a d-b c)}+6 x^2 (a+b x)^2}{12 b d \sqrt{a+b x} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sqrt[a + b*x])/(c + d*x)^(5/2),x]

[Out]

(6*x^2*(a + b*x)^2 + (2*c*(a + b*x)^2*(3*a*d*(c + 2*d*x) - 7*b*c*(5*c + 6*d*x)))/(d^2*(-(b*c) + a*d)) + (3*(35
*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*(c + d*x)^2*(-(Sqrt[d]*Sqrt[b*c - a*d]*(a + b*x)*Sqrt[(b*(c + d*x))/(b*c - a*
d)]) + (b*c - a*d)*Sqrt[a + b*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]))/(d^(7/2)*(b*c - a*d)^(3/2)
*Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(12*b*d*Sqrt[a + b*x]*(c + d*x)^(3/2))

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Maple [B]  time = 0.026, size = 986, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x+a)^(1/2)/(d*x+c)^(5/2),x)

[Out]

-1/24*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^3*d^5+27*ln(1/2*(2*
b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^2*b*c*d^4-135*ln(1/2*(2*b*d*x+2*((b*x+
a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a*b^2*c^2*d^3+105*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^
(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*b^3*c^3*d^2-12*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^3*a*b*d^4+12*
(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^3*b^2*c*d^3+6*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+
b*c)/(b*d)^(1/2))*x*a^3*c*d^4+54*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x
*a^2*b*c^2*d^3-270*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a*b^2*c^3*d^2
+210*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^3*c^4*d-6*(b*d)^(1/2)*((b
*x+a)*(d*x+c))^(1/2)*x^2*a^2*d^4+48*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a*b*c*d^3-42*(b*d)^(1/2)*((b*x+a)*
(d*x+c))^(1/2)*x^2*b^2*c^2*d^2+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a
^3*c^2*d^3+27*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c^3*d^2-135*ln
(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c^4*d+105*ln(1/2*(2*b*d*x+2*((
b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^5-12*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^2*c
*d^3+276*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b*c^2*d^2-280*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*b^2*c^3*d
-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c^2*d^2+200*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c^3*d-210*(b*d)
^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^4)*(b*x+a)^(1/2)/(a*d-b*c)/(b*d)^(1/2)/b/((b*x+a)*(d*x+c))^(1/2)/d^4/(d*x
+c)^(3/2)

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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(x^3*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.74666, size = 1827, normalized size = 8.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(35*b^3*c^5 - 45*a*b^2*c^4*d + 9*a^2*b*c^3*d^2 + a^3*c^2*d^3 + (35*b^3*c^3*d^2 - 45*a*b^2*c^2*d^3 +
9*a^2*b*c*d^4 + a^3*d^5)*x^2 + 2*(35*b^3*c^4*d - 45*a*b^2*c^3*d^2 + 9*a^2*b*c^2*d^3 + a^3*c*d^4)*x)*sqrt(b*d)*
log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x +
c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(105*b^3*c^4*d - 100*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 - 6*(b^3*c*d^4 - a*b^2
*d^5)*x^3 + 3*(7*b^3*c^2*d^3 - 8*a*b^2*c*d^4 + a^2*b*d^5)*x^2 + 2*(70*b^3*c^3*d^2 - 69*a*b^2*c^2*d^3 + 3*a^2*b
*c*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*c^3*d^5 - a*b^2*c^2*d^6 + (b^3*c*d^7 - a*b^2*d^8)*x^2 + 2*(b^3*c^
2*d^6 - a*b^2*c*d^7)*x), -1/24*(3*(35*b^3*c^5 - 45*a*b^2*c^4*d + 9*a^2*b*c^3*d^2 + a^3*c^2*d^3 + (35*b^3*c^3*d
^2 - 45*a*b^2*c^2*d^3 + 9*a^2*b*c*d^4 + a^3*d^5)*x^2 + 2*(35*b^3*c^4*d - 45*a*b^2*c^3*d^2 + 9*a^2*b*c^2*d^3 +
a^3*c*d^4)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2
+ a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(105*b^3*c^4*d - 100*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 - 6*(b^3*c*d^4 -
a*b^2*d^5)*x^3 + 3*(7*b^3*c^2*d^3 - 8*a*b^2*c*d^4 + a^2*b*d^5)*x^2 + 2*(70*b^3*c^3*d^2 - 69*a*b^2*c^2*d^3 + 3*
a^2*b*c*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*c^3*d^5 - a*b^2*c^2*d^6 + (b^3*c*d^7 - a*b^2*d^8)*x^2 + 2*(b
^3*c^2*d^6 - a*b^2*c*d^7)*x)]

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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(x**3*(b*x+a)**(1/2)/(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.40915, size = 549, normalized size = 2.52 \begin{align*} \frac{{\left ({\left (3 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (b^{5} c d^{6}{\left | b \right |} - a b^{4} d^{7}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{6} c d^{7} - a b^{5} d^{8}} - \frac{7 \, b^{6} c^{2} d^{5}{\left | b \right |} - 2 \, a b^{5} c d^{6}{\left | b \right |} - 5 \, a^{2} b^{4} d^{7}{\left | b \right |}}{b^{6} c d^{7} - a b^{5} d^{8}}\right )} - \frac{4 \,{\left (35 \, b^{7} c^{3} d^{4}{\left | b \right |} - 45 \, a b^{6} c^{2} d^{5}{\left | b \right |} + 9 \, a^{2} b^{5} c d^{6}{\left | b \right |} + 3 \, a^{3} b^{4} d^{7}{\left | b \right |}\right )}}{b^{6} c d^{7} - a b^{5} d^{8}}\right )}{\left (b x + a\right )} - \frac{3 \,{\left (35 \, b^{8} c^{4} d^{3}{\left | b \right |} - 80 \, a b^{7} c^{3} d^{4}{\left | b \right |} + 54 \, a^{2} b^{6} c^{2} d^{5}{\left | b \right |} - 8 \, a^{3} b^{5} c d^{6}{\left | b \right |} - a^{4} b^{4} d^{7}{\left | b \right |}\right )}}{b^{6} c d^{7} - a b^{5} d^{8}}\right )} \sqrt{b x + a}}{12 \,{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} - \frac{{\left (35 \, b^{2} c^{2}{\left | b \right |} - 10 \, a b c d{\left | b \right |} - a^{2} d^{2}{\left | b \right |}\right )} \log \left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt{b d} b^{2} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/12*((3*(b*x + a)*(2*(b^5*c*d^6*abs(b) - a*b^4*d^7*abs(b))*(b*x + a)/(b^6*c*d^7 - a*b^5*d^8) - (7*b^6*c^2*d^5
*abs(b) - 2*a*b^5*c*d^6*abs(b) - 5*a^2*b^4*d^7*abs(b))/(b^6*c*d^7 - a*b^5*d^8)) - 4*(35*b^7*c^3*d^4*abs(b) - 4
5*a*b^6*c^2*d^5*abs(b) + 9*a^2*b^5*c*d^6*abs(b) + 3*a^3*b^4*d^7*abs(b))/(b^6*c*d^7 - a*b^5*d^8))*(b*x + a) - 3
*(35*b^8*c^4*d^3*abs(b) - 80*a*b^7*c^3*d^4*abs(b) + 54*a^2*b^6*c^2*d^5*abs(b) - 8*a^3*b^5*c*d^6*abs(b) - a^4*b
^4*d^7*abs(b))/(b^6*c*d^7 - a*b^5*d^8))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(35*b^2*c^2*
abs(b) - 10*a*b*c*d*abs(b) - a^2*d^2*abs(b))*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a
*b*d)))/(sqrt(b*d)*b^2*d^4)