3.748 \(\int \frac{x^4}{\sqrt{a+b x} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=258 \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (-2 b d x \left (3 a^2 d^2-46 a b c d+35 b^2 c^2\right )+15 a^2 b c d^2+9 a^3 d^3-145 a b^2 c^2 d+105 b^3 c^3\right )}{12 b^2 d^4 (b c-a d)^2}+\frac{\left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{9/2}}-\frac{2 c x^2 \sqrt{a+b x} (7 b c-9 a d)}{3 d^2 \sqrt{c+d x} (b c-a d)^2}-\frac{2 c x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2} (b c-a d)} \]

[Out]

(-2*c*x^3*Sqrt[a + b*x])/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) - (2*c*(7*b*c - 9*a*d)*x^2*Sqrt[a + b*x])/(3*d^2*(b
*c - a*d)^2*Sqrt[c + d*x]) - (Sqrt[a + b*x]*Sqrt[c + d*x]*(105*b^3*c^3 - 145*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 9*
a^3*d^3 - 2*b*d*(35*b^2*c^2 - 46*a*b*c*d + 3*a^2*d^2)*x))/(12*b^2*d^4*(b*c - a*d)^2) + ((35*b^2*c^2 + 10*a*b*c
*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.220917, antiderivative size = 258, 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 = {98, 150, 147, 63, 217, 206} \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (-2 b d x \left (3 a^2 d^2-46 a b c d+35 b^2 c^2\right )+15 a^2 b c d^2+9 a^3 d^3-145 a b^2 c^2 d+105 b^3 c^3\right )}{12 b^2 d^4 (b c-a d)^2}+\frac{\left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{9/2}}-\frac{2 c x^2 \sqrt{a+b x} (7 b c-9 a d)}{3 d^2 \sqrt{c+d x} (b c-a d)^2}-\frac{2 c x^3 \sqrt{a+b x}}{3 d (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

(-2*c*x^3*Sqrt[a + b*x])/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) - (2*c*(7*b*c - 9*a*d)*x^2*Sqrt[a + b*x])/(3*d^2*(b
*c - a*d)^2*Sqrt[c + d*x]) - (Sqrt[a + b*x]*Sqrt[c + d*x]*(105*b^3*c^3 - 145*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 9*
a^3*d^3 - 2*b*d*(35*b^2*c^2 - 46*a*b*c*d + 3*a^2*d^2)*x))/(12*b^2*d^4*(b*c - a*d)^2) + ((35*b^2*c^2 + 10*a*b*c
*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(9/2))

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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^4}{\sqrt{a+b x} (c+d x)^{5/2}} \, dx &=-\frac{2 c x^3 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac{2 \int \frac{x^2 \left (3 a c+\frac{1}{2} (7 b c-3 a d) x\right )}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 d (b c-a d)}\\ &=-\frac{2 c x^3 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac{2 c (7 b c-9 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{4 \int \frac{x \left (-a c (7 b c-9 a d)+\frac{1}{4} \left (-35 b^2 c^2+46 a b c d-3 a^2 d^2\right ) x\right )}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 d^2 (b c-a d)^2}\\ &=-\frac{2 c x^3 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac{2 c (7 b c-9 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^3 c^3-145 a b^2 c^2 d+15 a^2 b c d^2+9 a^3 d^3-2 b d \left (35 b^2 c^2-46 a b c d+3 a^2 d^2\right ) x\right )}{12 b^2 d^4 (b c-a d)^2}+\frac{\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b^2 d^4}\\ &=-\frac{2 c x^3 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac{2 c (7 b c-9 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^3 c^3-145 a b^2 c^2 d+15 a^2 b c d^2+9 a^3 d^3-2 b d \left (35 b^2 c^2-46 a b c d+3 a^2 d^2\right ) x\right )}{12 b^2 d^4 (b c-a d)^2}+\frac{\left (35 b^2 c^2+10 a b c d+3 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^3 d^4}\\ &=-\frac{2 c x^3 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac{2 c (7 b c-9 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^3 c^3-145 a b^2 c^2 d+15 a^2 b c d^2+9 a^3 d^3-2 b d \left (35 b^2 c^2-46 a b c d+3 a^2 d^2\right ) x\right )}{12 b^2 d^4 (b c-a d)^2}+\frac{\left (35 b^2 c^2+10 a b c d+3 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^3 d^4}\\ &=-\frac{2 c x^3 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac{2 c (7 b c-9 a d) x^2 \sqrt{a+b x}}{3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (105 b^3 c^3-145 a b^2 c^2 d+15 a^2 b c d^2+9 a^3 d^3-2 b d \left (35 b^2 c^2-46 a b c d+3 a^2 d^2\right ) x\right )}{12 b^2 d^4 (b c-a d)^2}+\frac{\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.951259, size = 367, normalized size = 1.42 \[ \frac{-3 (c+d x) (b c-a d) (3 a d+7 b c) \left (-d \sqrt{b c-a d} \left (-a^2 d^2-2 a b c d+3 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )+2 b^2 c^2 d^{3/2} \sqrt{a+b x}+b d^{3/2} \sqrt{a+b x} (c+d x) (b c-a d)\right )+6 b^2 d^{9/2} x^3 \sqrt{a+b x} (b c-a d)^2-2 b c d (3 a d-7 b c) \left (b c^2 \sqrt{d} \sqrt{a+b x} (b c-a d)-(c+d x) \left (2 b c \sqrt{d} \sqrt{a+b x} (2 b c-3 a d)-3 (b c-a d)^{5/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )\right )\right )}{12 b^3 d^{11/2} (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

(6*b^2*d^(9/2)*(b*c - a*d)^2*x^3*Sqrt[a + b*x] - 3*(b*c - a*d)*(7*b*c + 3*a*d)*(c + d*x)*(2*b^2*c^2*d^(3/2)*Sq
rt[a + b*x] + b*d^(3/2)*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x) - d*Sqrt[b*c - a*d]*(3*b^2*c^2 - 2*a*b*c*d - a^2*d
^2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]) - 2*b*c*d*(-7*b*c + 3*a*
d)*(b*c^2*Sqrt[d]*(b*c - a*d)*Sqrt[a + b*x] - (c + d*x)*(2*b*c*Sqrt[d]*(2*b*c - 3*a*d)*Sqrt[a + b*x] - 3*(b*c
- a*d)^(5/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])))/(12*b^3*d^(11
/2)*(b*c - a*d)^2*(c + d*x)^(3/2))

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Maple [B]  time = 0.03, size = 1287, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(d*x+c)^(5/2)/(b*x+a)^(1/2),x)

[Out]

1/24*(b*x+a)^(1/2)*(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^4*c^
4*d^2+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^4*c^6-210*(b*d)^(1/2)*
((b*x+a)*(d*x+c))^(1/2)*b^3*c^5+9*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^4*d^6+9*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^4*c^2*d^4+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^4*c^5*d+12*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*b*c^3*d^3+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))*a^2*b^2*c^4*d^2-180*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^3*c^5*d-18*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^3*d^5-18*(b*d)^(1/2)*((b*x+a)*(
d*x+c))^(1/2)*a^3*c^2*d^3+18*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^4
*c*d^5+12*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*b*c*d^5+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^2*a^2*b^2*c^2*d^4-180*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^3*c^3*d^3-42*(b*d)^(1/2)*((b*x+a)*(d*x+c)
)^(1/2)*x^2*b^3*c^3*d^2+24*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*b
*c^2*d^4+108*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^2*c^3*d^3-360
*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^3*c^4*d^2-280*(b*d)^(1/2)*(
(b*x+a)*(d*x+c))^(1/2)*x*b^3*c^4*d-30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^3*d^2+290*(b*d)^(1/2)*((b*x+
a)*(d*x+c))^(1/2)*a*b^2*c^4*d+12*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^3*a^2*b*d^5+12*(b*d)^(1/2)*((b*x+a)*(d*
x+c))^(1/2)*x^3*b^3*c^2*d^3-36*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^3*c*d^4-24*(b*d)^(1/2)*((b*x+a)*(d*x+c)
)^(1/2)*x^3*a*b^2*c*d^4-48*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^2*b*c^2*d^3+396*(b*d)^(1/2)*((b*x+a)*(d*x+c
))^(1/2)*x*a*b^2*c^3*d^2-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^2*b*c*d^4+66*(b*d)^(1/2)*((b*x+a)*(d*x+c)
)^(1/2)*x^2*a*b^2*c^2*d^3)/(a*d-b*c)^2/(b*d)^(1/2)/b^2/((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^4/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.81778, size = 2398, normalized size = 9.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*(35*b^4*c^6 - 60*a*b^3*c^5*d + 18*a^2*b^2*c^4*d^2 + 4*a^3*b*c^3*d^3 + 3*a^4*c^2*d^4 + (35*b^4*c^4*d^2
- 60*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 + 4*a^3*b*c*d^5 + 3*a^4*d^6)*x^2 + 2*(35*b^4*c^5*d - 60*a*b^3*c^4*d^2
+ 18*a^2*b^2*c^3*d^3 + 4*a^3*b*c^2*d^4 + 3*a^4*c*d^5)*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^
4*c^5*d - 145*a*b^3*c^4*d^2 + 15*a^2*b^2*c^3*d^3 + 9*a^3*b*c^2*d^4 - 6*(b^4*c^2*d^4 - 2*a*b^3*c*d^5 + a^2*b^2*
d^6)*x^3 + 3*(7*b^4*c^3*d^3 - 11*a*b^3*c^2*d^4 + a^2*b^2*c*d^5 + 3*a^3*b*d^6)*x^2 + 2*(70*b^4*c^4*d^2 - 99*a*b
^3*c^3*d^3 + 12*a^2*b^2*c^2*d^4 + 9*a^3*b*c*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*c^4*d^5 - 2*a*b^4*c^3*d^
6 + a^2*b^3*c^2*d^7 + (b^5*c^2*d^7 - 2*a*b^4*c*d^8 + a^2*b^3*d^9)*x^2 + 2*(b^5*c^3*d^6 - 2*a*b^4*c^2*d^7 + a^2
*b^3*c*d^8)*x), -1/24*(3*(35*b^4*c^6 - 60*a*b^3*c^5*d + 18*a^2*b^2*c^4*d^2 + 4*a^3*b*c^3*d^3 + 3*a^4*c^2*d^4 +
(35*b^4*c^4*d^2 - 60*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 + 4*a^3*b*c*d^5 + 3*a^4*d^6)*x^2 + 2*(35*b^4*c^5*d -
60*a*b^3*c^4*d^2 + 18*a^2*b^2*c^3*d^3 + 4*a^3*b*c^2*d^4 + 3*a^4*c*d^5)*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^4*c
^5*d - 145*a*b^3*c^4*d^2 + 15*a^2*b^2*c^3*d^3 + 9*a^3*b*c^2*d^4 - 6*(b^4*c^2*d^4 - 2*a*b^3*c*d^5 + a^2*b^2*d^6
)*x^3 + 3*(7*b^4*c^3*d^3 - 11*a*b^3*c^2*d^4 + a^2*b^2*c*d^5 + 3*a^3*b*d^6)*x^2 + 2*(70*b^4*c^4*d^2 - 99*a*b^3*
c^3*d^3 + 12*a^2*b^2*c^2*d^4 + 9*a^3*b*c*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*c^4*d^5 - 2*a*b^4*c^3*d^6 +
a^2*b^3*c^2*d^7 + (b^5*c^2*d^7 - 2*a*b^4*c*d^8 + a^2*b^3*d^9)*x^2 + 2*(b^5*c^3*d^6 - 2*a*b^4*c^2*d^7 + a^2*b^
3*c*d^8)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(d*x+c)**(5/2)/(b*x+a)**(1/2),x)

[Out]

Integral(x**4/(sqrt(a + b*x)*(c + d*x)**(5/2)), x)

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Giac [B]  time = 1.9354, size = 686, normalized size = 2.66 \begin{align*} \frac{{\left ({\left (3 \,{\left (b x + a\right )}{\left (\frac{2 \,{\left (b^{7} c^{2} d^{6} - 2 \, a b^{6} c d^{7} + a^{2} b^{5} d^{8}\right )}{\left (b x + a\right )}}{b^{7} c^{2} d^{7}{\left | b \right |} - 2 \, a b^{6} c d^{8}{\left | b \right |} + a^{2} b^{5} d^{9}{\left | b \right |}} - \frac{7 \, b^{8} c^{3} d^{5} - 5 \, a b^{7} c^{2} d^{6} - 11 \, a^{2} b^{6} c d^{7} + 9 \, a^{3} b^{5} d^{8}}{b^{7} c^{2} d^{7}{\left | b \right |} - 2 \, a b^{6} c d^{8}{\left | b \right |} + a^{2} b^{5} d^{9}{\left | b \right |}}\right )} - \frac{4 \,{\left (35 \, b^{9} c^{4} d^{4} - 60 \, a b^{8} c^{3} d^{5} + 18 \, a^{2} b^{7} c^{2} d^{6} + 12 \, a^{3} b^{6} c d^{7} - 9 \, a^{4} b^{5} d^{8}\right )}}{b^{7} c^{2} d^{7}{\left | b \right |} - 2 \, a b^{6} c d^{8}{\left | b \right |} + a^{2} b^{5} d^{9}{\left | b \right |}}\right )}{\left (b x + a\right )} - \frac{3 \,{\left (35 \, b^{10} c^{5} d^{3} - 95 \, a b^{9} c^{4} d^{4} + 78 \, a^{2} b^{8} c^{3} d^{5} - 14 \, a^{3} b^{7} c^{2} d^{6} - 9 \, a^{4} b^{6} c d^{7} + 5 \, a^{5} b^{5} d^{8}\right )}}{b^{7} c^{2} d^{7}{\left | b \right |} - 2 \, a b^{6} c d^{8}{\left | b \right |} + a^{2} b^{5} d^{9}{\left | b \right |}}\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} + 10 \, a b c d + 3 \, a^{2} d^{2}\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 d^{4}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/12*((3*(b*x + a)*(2*(b^7*c^2*d^6 - 2*a*b^6*c*d^7 + a^2*b^5*d^8)*(b*x + a)/(b^7*c^2*d^7*abs(b) - 2*a*b^6*c*d^
8*abs(b) + a^2*b^5*d^9*abs(b)) - (7*b^8*c^3*d^5 - 5*a*b^7*c^2*d^6 - 11*a^2*b^6*c*d^7 + 9*a^3*b^5*d^8)/(b^7*c^2
*d^7*abs(b) - 2*a*b^6*c*d^8*abs(b) + a^2*b^5*d^9*abs(b))) - 4*(35*b^9*c^4*d^4 - 60*a*b^8*c^3*d^5 + 18*a^2*b^7*
c^2*d^6 + 12*a^3*b^6*c*d^7 - 9*a^4*b^5*d^8)/(b^7*c^2*d^7*abs(b) - 2*a*b^6*c*d^8*abs(b) + a^2*b^5*d^9*abs(b)))*
(b*x + a) - 3*(35*b^10*c^5*d^3 - 95*a*b^9*c^4*d^4 + 78*a^2*b^8*c^3*d^5 - 14*a^3*b^7*c^2*d^6 - 9*a^4*b^6*c*d^7
+ 5*a^5*b^5*d^8)/(b^7*c^2*d^7*abs(b) - 2*a*b^6*c*d^8*abs(b) + a^2*b^5*d^9*abs(b)))*sqrt(b*x + a)/(b^2*c + (b*x
+ a)*b*d - a*b*d)^(3/2) - 1/4*(35*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*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*d^4*abs(b))