\(\int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) [785]

Optimal result

Integrand size = 22, antiderivative size = 251 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt {c+d x}}-\frac {(5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} d^{7/2}} \]

[Out]

-(3*a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(7/2)+2*a*x^3/b/(-a*d+b*c)/(d*x+
c)^(3/2)/(b*x+a)^(1/2)-2/3*c*(3*a*d+b*c)*x^2*(b*x+a)^(1/2)/b/d/(-a*d+b*c)^2/(d*x+c)^(3/2)+1/3*(c*(-9*a^3*d^3+9
*a^2*b*c*d^2-31*a*b^2*c^2*d+15*b^3*c^3)+d*(-a*d+b*c)*(9*a^2*d^2-6*a*b*c*d+5*b^2*c^2)*x)*(b*x+a)^(1/2)/b^2/d^3/
(-a*d+b*c)^3/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {100, 155, 148, 65, 223, 212} \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{3 b^2 d^3 \sqrt {c+d x} (b c-a d)^3}-\frac {(3 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} d^{7/2}}+\frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 b d (c+d x)^{3/2} (b c-a d)^2} \]

[In]

Int[x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]

[Out]

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (2*c*(b*c + 3*a*d)*x^2*Sqrt[a + b*x])/(3*b*d*(b*c -
a*d)^2*(c + d*x)^(3/2)) + (Sqrt[a + b*x]*(c*(15*b^3*c^3 - 31*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 9*a^3*d^3) + d*(b*c
- a*d)*(5*b^2*c^2 - 6*a*b*c*d + 9*a^2*d^2)*x))/(3*b^2*d^3*(b*c - a*d)^3*Sqrt[c + d*x]) - ((5*b*c + 3*a*d)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(5/2)*d^(7/2))

Rule 65

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 100

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 148

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

Rule 155

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 212

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

Rule 223

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 \int \frac {x^2 \left (3 a c+\frac {1}{2} (-b c+3 a d) x\right )}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx}{b (b c-a d)} \\ & = \frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {4 \int \frac {x \left (a c (b c+3 a d)+\frac {1}{4} \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 b d (b c-a d)^2} \\ & = \frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt {c+d x}}-\frac {(5 b c+3 a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 b^2 d^3} \\ & = \frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt {c+d x}}-\frac {(5 b c+3 a d) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{b^3 d^3} \\ & = \frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt {c+d x}}-\frac {(5 b c+3 a d) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{b^3 d^3} \\ & = \frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt {c+d x}}-\frac {(5 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {9 a^4 d^3 (c+d x)^2+3 a^3 b d^2 (-3 c+d x) (c+d x)^2-b^4 c^3 x \left (15 c^2+20 c d x+3 d^2 x^2\right )+a^2 b^2 c d \left (31 c^3+33 c^2 d x-9 c d^2 x^2-9 d^3 x^3\right )+a b^3 c^2 \left (-15 c^3+11 c^2 d x+39 c d^2 x^2+9 d^3 x^3\right )}{3 b^2 d^3 (-b c+a d)^3 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {(5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2} d^{7/2}} \]

[In]

Integrate[x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]

[Out]

(9*a^4*d^3*(c + d*x)^2 + 3*a^3*b*d^2*(-3*c + d*x)*(c + d*x)^2 - b^4*c^3*x*(15*c^2 + 20*c*d*x + 3*d^2*x^2) + a^
2*b^2*c*d*(31*c^3 + 33*c^2*d*x - 9*c*d^2*x^2 - 9*d^3*x^3) + a*b^3*c^2*(-15*c^3 + 11*c^2*d*x + 39*c*d^2*x^2 + 9
*d^3*x^3))/(3*b^2*d^3*(-(b*c) + a*d)^3*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((5*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt
[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(b^(5/2)*d^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1713\) vs. \(2(223)=446\).

Time = 0.59 (sec) , antiderivative size = 1714, normalized size of antiderivative = 6.83

method	result	size
default	\(\text {Expression too large to display}\)	\(1714\)

[In]

int(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(-30*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^5*c^5*d*x^2-18*a^4*d^5
*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*b^4*c^5*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-18*a^4*c^2*d^3*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-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))*
a^3*b^2*c^4*d^2-15*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^4*c^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^5*d^6*x^2-15*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^5*c^6*x+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^5*c^2*d^4-62*a^2*b^2*c^4*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-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^2*c*d^5*x^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))*a^2*b^3*c^2*d^4*x^3+36*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^4*c^3*d^3*x^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))*a^4*b*c*d^5*x^2-42*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^2*c^2*d^4*x^2+57*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^4*c^4*d^2*x^2-15*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*b*c^2*d
^4*x-48*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^2*c^3*d^3*x+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^3*c^4*d^2*x+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))*a*b^4*c^5*d*x+30*a*b^3*c^5*((b*x+a)*(d*x+c))^(1/2)*(b
*d)^(1/2)+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*b*d^6*x^3-15*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^5*c^4*d^2*x^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))*a^5*c*d^5*x-6*a^3*b*d^5*x^3*((b*x+a)*(d*x+c))^(1/2)*(b
*d)^(1/2)+6*b^4*c^3*d^2*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+40*b^4*c^4*d*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)
^(1/2)-36*a^4*c*d^4*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+18*a^3*b*c^3*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)
-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^4*b*c^3*d^3+36*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^3*c^5*d+18*a^2*b^2*c*d^4*x^3*((b*x+a)*(d*
x+c))^(1/2)*(b*d)^(1/2)-18*a*b^3*c^2*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+6*a^3*b*c*d^4*x^2*((b*x+a)*(d
*x+c))^(1/2)*(b*d)^(1/2)+18*a^2*b^2*c^2*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-78*a*b^3*c^3*d^2*x^2*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*a^3*b*c^2*d^3*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-66*a^2*b^2*c^3*d^2*x*((b
*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-22*a*b^3*c^4*d*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2)/(a*d-b*c)^3
/((b*x+a)*(d*x+c))^(1/2)/d^3/b^2/(d*x+c)^(3/2)/(b*x+a)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 807 vs. \(2 (223) = 446\).

Time = 0.58 (sec) , antiderivative size = 1628, normalized size of antiderivative = 6.49 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(5*a*b^4*c^6 - 12*a^2*b^3*c^5*d + 6*a^3*b^2*c^4*d^2 + 4*a^4*b*c^3*d^3 - 3*a^5*c^2*d^4 + (5*b^5*c^4*d^
2 - 12*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 + 4*a^3*b^2*c*d^5 - 3*a^4*b*d^6)*x^3 + (10*b^5*c^5*d - 19*a*b^4*c^4*d
^2 + 14*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 - 3*a^5*d^6)*x^2 + (5*b^5*c^6 - 2*a*b^4*c^5*d - 18*a^2*b^3*c^4*d^2 + 1
6*a^3*b^2*c^3*d^3 + 5*a^4*b*c^2*d^4 - 6*a^5*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*(15*a*b^4*c
^5*d - 31*a^2*b^3*c^4*d^2 + 9*a^3*b^2*c^3*d^3 - 9*a^4*b*c^2*d^4 + 3*(b^5*c^3*d^3 - 3*a*b^4*c^2*d^4 + 3*a^2*b^3
*c*d^5 - a^3*b^2*d^6)*x^3 + (20*b^5*c^4*d^2 - 39*a*b^4*c^3*d^3 + 9*a^2*b^3*c^2*d^4 + 3*a^3*b^2*c*d^5 - 9*a^4*b
*d^6)*x^2 + (15*b^5*c^5*d - 11*a*b^4*c^4*d^2 - 33*a^2*b^3*c^3*d^3 + 15*a^3*b^2*c^2*d^4 - 18*a^4*b*c*d^5)*x)*sq
rt(b*x + a)*sqrt(d*x + c))/(a*b^6*c^5*d^4 - 3*a^2*b^5*c^4*d^5 + 3*a^3*b^4*c^3*d^6 - a^4*b^3*c^2*d^7 + (b^7*c^3
*d^6 - 3*a*b^6*c^2*d^7 + 3*a^2*b^5*c*d^8 - a^3*b^4*d^9)*x^3 + (2*b^7*c^4*d^5 - 5*a*b^6*c^3*d^6 + 3*a^2*b^5*c^2
*d^7 + a^3*b^4*c*d^8 - a^4*b^3*d^9)*x^2 + (b^7*c^5*d^4 - a*b^6*c^4*d^5 - 3*a^2*b^5*c^3*d^6 + 5*a^3*b^4*c^2*d^7
- 2*a^4*b^3*c*d^8)*x), 1/6*(3*(5*a*b^4*c^6 - 12*a^2*b^3*c^5*d + 6*a^3*b^2*c^4*d^2 + 4*a^4*b*c^3*d^3 - 3*a^5*c
^2*d^4 + (5*b^5*c^4*d^2 - 12*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 + 4*a^3*b^2*c*d^5 - 3*a^4*b*d^6)*x^3 + (10*b^5*
c^5*d - 19*a*b^4*c^4*d^2 + 14*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 - 3*a^5*d^6)*x^2 + (5*b^5*c^6 - 2*a*b^4*c^5*d -
18*a^2*b^3*c^4*d^2 + 16*a^3*b^2*c^3*d^3 + 5*a^4*b*c^2*d^4 - 6*a^5*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*(15*a*b^
4*c^5*d - 31*a^2*b^3*c^4*d^2 + 9*a^3*b^2*c^3*d^3 - 9*a^4*b*c^2*d^4 + 3*(b^5*c^3*d^3 - 3*a*b^4*c^2*d^4 + 3*a^2*
b^3*c*d^5 - a^3*b^2*d^6)*x^3 + (20*b^5*c^4*d^2 - 39*a*b^4*c^3*d^3 + 9*a^2*b^3*c^2*d^4 + 3*a^3*b^2*c*d^5 - 9*a^
4*b*d^6)*x^2 + (15*b^5*c^5*d - 11*a*b^4*c^4*d^2 - 33*a^2*b^3*c^3*d^3 + 15*a^3*b^2*c^2*d^4 - 18*a^4*b*c*d^5)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^6*c^5*d^4 - 3*a^2*b^5*c^4*d^5 + 3*a^3*b^4*c^3*d^6 - a^4*b^3*c^2*d^7 + (b^7*
c^3*d^6 - 3*a*b^6*c^2*d^7 + 3*a^2*b^5*c*d^8 - a^3*b^4*d^9)*x^3 + (2*b^7*c^4*d^5 - 5*a*b^6*c^3*d^6 + 3*a^2*b^5*
c^2*d^7 + a^3*b^4*c*d^8 - a^4*b^3*d^9)*x^2 + (b^7*c^5*d^4 - a*b^6*c^4*d^5 - 3*a^2*b^5*c^3*d^6 + 5*a^3*b^4*c^2*
d^7 - 2*a^4*b^3*c*d^8)*x)]

Sympy [F]

\[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {x^{4}}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**4/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)

[Out]

Integral(x**4/((a + b*x)**(3/2)*(c + d*x)**(5/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 756 vs. \(2 (223) = 446\).

Time = 0.50 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.01 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {4 \, a^{4} d}{{\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} + \frac {{\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (b^{8} c^{5} d^{4} {\left | b \right |} - 5 \, a b^{7} c^{4} d^{5} {\left | b \right |} + 10 \, a^{2} b^{6} c^{3} d^{6} {\left | b \right |} - 10 \, a^{3} b^{5} c^{2} d^{7} {\left | b \right |} + 5 \, a^{4} b^{4} c d^{8} {\left | b \right |} - a^{5} b^{3} d^{9} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{10} c^{5} d^{5} - 5 \, a b^{9} c^{4} d^{6} + 10 \, a^{2} b^{8} c^{3} d^{7} - 10 \, a^{3} b^{7} c^{2} d^{8} + 5 \, a^{4} b^{6} c d^{9} - a^{5} b^{5} d^{10}} + \frac {2 \, {\left (10 \, b^{9} c^{6} d^{3} {\left | b \right |} - 44 \, a b^{8} c^{5} d^{4} {\left | b \right |} + 76 \, a^{2} b^{7} c^{4} d^{5} {\left | b \right |} - 72 \, a^{3} b^{6} c^{3} d^{6} {\left | b \right |} + 45 \, a^{4} b^{5} c^{2} d^{7} {\left | b \right |} - 18 \, a^{5} b^{4} c d^{8} {\left | b \right |} + 3 \, a^{6} b^{3} d^{9} {\left | b \right |}\right )}}{b^{10} c^{5} d^{5} - 5 \, a b^{9} c^{4} d^{6} + 10 \, a^{2} b^{8} c^{3} d^{7} - 10 \, a^{3} b^{7} c^{2} d^{8} + 5 \, a^{4} b^{6} c d^{9} - a^{5} b^{5} d^{10}}\right )} + \frac {3 \, {\left (5 \, b^{10} c^{7} d^{2} {\left | b \right |} - 27 \, a b^{9} c^{6} d^{3} {\left | b \right |} + 57 \, a^{2} b^{8} c^{5} d^{4} {\left | b \right |} - 63 \, a^{3} b^{7} c^{4} d^{5} {\left | b \right |} + 43 \, a^{4} b^{6} c^{3} d^{6} {\left | b \right |} - 21 \, a^{5} b^{5} c^{2} d^{7} {\left | b \right |} + 7 \, a^{6} b^{4} c d^{8} {\left | b \right |} - a^{7} b^{3} d^{9} {\left | b \right |}\right )}}{b^{10} c^{5} d^{5} - 5 \, a b^{9} c^{4} d^{6} + 10 \, a^{2} b^{8} c^{3} d^{7} - 10 \, a^{3} b^{7} c^{2} d^{8} + 5 \, a^{4} b^{6} c d^{9} - a^{5} b^{5} d^{10}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, b c + 3 \, a d\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 )}^{2}\right )}{2 \, \sqrt {b d} b d^{3} {\left | b \right |}} \]

[In]

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-4*a^4*d/((sqrt(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*d^2*abs(b))*(b^2*c - a*b*d -
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)) + 1/3*((b*x + a)*(3*(b^8*c^5*d^4*abs(b) -
5*a*b^7*c^4*d^5*abs(b) + 10*a^2*b^6*c^3*d^6*abs(b) - 10*a^3*b^5*c^2*d^7*abs(b) + 5*a^4*b^4*c*d^8*abs(b) - a^5*
b^3*d^9*abs(b))*(b*x + a)/(b^10*c^5*d^5 - 5*a*b^9*c^4*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^
6*c*d^9 - a^5*b^5*d^10) + 2*(10*b^9*c^6*d^3*abs(b) - 44*a*b^8*c^5*d^4*abs(b) + 76*a^2*b^7*c^4*d^5*abs(b) - 72*
a^3*b^6*c^3*d^6*abs(b) + 45*a^4*b^5*c^2*d^7*abs(b) - 18*a^5*b^4*c*d^8*abs(b) + 3*a^6*b^3*d^9*abs(b))/(b^10*c^5
*d^5 - 5*a*b^9*c^4*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*d^10)) + 3*(5*b^1
0*c^7*d^2*abs(b) - 27*a*b^9*c^6*d^3*abs(b) + 57*a^2*b^8*c^5*d^4*abs(b) - 63*a^3*b^7*c^4*d^5*abs(b) + 43*a^4*b^
6*c^3*d^6*abs(b) - 21*a^5*b^5*c^2*d^7*abs(b) + 7*a^6*b^4*c*d^8*abs(b) - a^7*b^3*d^9*abs(b))/(b^10*c^5*d^5 - 5*
a*b^9*c^4*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*d^10))*sqrt(b*x + a)/(b^2*
c + (b*x + a)*b*d - a*b*d)^(3/2) + 1/2*(5*b*c + 3*a*d)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b
*d - a*b*d))^2)/(sqrt(b*d)*b*d^3*abs(b))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {x^4}{{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

int(x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x)

[Out]

int(x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)), x)