\(\int \frac {x^4}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx\) [356]

Optimal result

Integrand size = 22, antiderivative size = 218 \[ \int \frac {x^4}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {2 c^4 \sqrt {a+b x}}{3 d^4 (b c-a d) (c+d x)^{3/2}}-\frac {4 c^3 (5 b c-6 a d) \sqrt {a+b x}}{3 d^4 (b c-a d)^2 \sqrt {c+d x}}-\frac {(13 b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2 d^4}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d^4}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{9/2}} \] Output:

2/3*c^4*(b*x+a)^(1/2)/d^4/(-a*d+b*c)/(d*x+c)^(3/2)-4/3*c^3*(-6*a*d+5*b*c)* 
(b*x+a)^(1/2)/d^4/(-a*d+b*c)^2/(d*x+c)^(1/2)-1/4*(3*a*d+13*b*c)*(b*x+a)^(1 
/2)*(d*x+c)^(1/2)/b^2/d^4+1/2*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b/d^4+1/4*(3*a^2 
*d^2+10*a*b*c*d+35*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^ 
(1/2))/b^(5/2)/d^(9/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.01 \[ \int \frac {x^4}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx=-\frac {\sqrt {a+b x} \left (9 a^3 d^3 (c+d x)^2+3 a^2 b d^2 (5 c-2 d x) (c+d x)^2-a b^2 c d \left (145 c^3+198 c^2 d x+33 c d^2 x^2-12 d^3 x^3\right )+b^3 c^2 \left (105 c^3+140 c^2 d x+21 c d^2 x^2-6 d^3 x^3\right )\right )}{12 b^2 d^4 (b c-a d)^2 (c+d x)^{3/2}}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{9/2}} \] Input:

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

Output:

-1/12*(Sqrt[a + b*x]*(9*a^3*d^3*(c + d*x)^2 + 3*a^2*b*d^2*(5*c - 2*d*x)*(c 
 + d*x)^2 - a*b^2*c*d*(145*c^3 + 198*c^2*d*x + 33*c*d^2*x^2 - 12*d^3*x^3) 
+ b^3*c^2*(105*c^3 + 140*c^2*d*x + 21*c*d^2*x^2 - 6*d^3*x^3)))/(b^2*d^4*(b 
*c - a*d)^2*(c + d*x)^(3/2)) + ((35*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*ArcT 
anh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(9/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {109, 27, 167, 27, 164, 66, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {2 \int \frac {x^2 (6 a c+(7 b c-3 a d) x)}{2 \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {x^2 (6 a c+(7 b c-3 a d) x)}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 167

\(\displaystyle \frac {-\frac {2 \int -\frac {x \left (4 a c (7 b c-9 a d)+\left (35 b^2 c^2-46 a b d c+3 a^2 d^2\right ) x\right )}{2 \sqrt {a+b x} \sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x} (7 b c-9 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {x \left (4 a c (7 b c-9 a d)+\left (35 b^2 c^2-46 a b d c+3 a^2 d^2\right ) x\right )}{\sqrt {a+b x} \sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x} (7 b c-9 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {\frac {\frac {3 (b c-a d)^2 \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{8 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (9 a^3 d^3-2 b d x \left (3 a^2 d^2-46 a b c d+35 b^2 c^2\right )+15 a^2 b c d^2-145 a b^2 c^2 d+105 b^3 c^3\right )}{4 b^2 d^2}}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x} (7 b c-9 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {\frac {3 (b c-a d)^2 \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{4 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (9 a^3 d^3-2 b d x \left (3 a^2 d^2-46 a b c d+35 b^2 c^2\right )+15 a^2 b c d^2-145 a b^2 c^2 d+105 b^3 c^3\right )}{4 b^2 d^2}}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x} (7 b c-9 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {3 (b c-a d)^2 \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (9 a^3 d^3-2 b d x \left (3 a^2 d^2-46 a b c d+35 b^2 c^2\right )+15 a^2 b c d^2-145 a b^2 c^2 d+105 b^3 c^3\right )}{4 b^2 d^2}}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x} (7 b c-9 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d (b c-a d)}-\frac {2 c x^3 \sqrt {a+b x}}{3 d (c+d x)^{3/2} (b c-a d)}\)

Input:

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

Output:

(-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])/(d*(b*c - a*d)*Sqrt[c + d*x]) + (-1/4*(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))/(b^2*d^2) + (3*( 
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^(5/2)))/(d*(b*c - a*d)))/( 
3*d*(b*c - a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 66

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

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> 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] + Simp[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 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> 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] + Simp[(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 167

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

rule 221

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1286\) vs. \(2(180)=360\).

Time = 0.28 (sec) , antiderivative size = 1287, normalized size of antiderivative = 5.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1287\)

Input:

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

Output:

1/24*(b*x+a)^(1/2)*(396*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^3*d^2* 
x-24*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a*b^2*c*d^4*x^3-36*((b*x+a)*(d*x+ 
c))^(1/2)*(d*b)^(1/2)*a^3*c*d^4*x+12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^( 
1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b*c*d^5*x^2+54*ln(1/2*(2*b*d*x+ 
2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*c^2*d^ 
4*x^2-180*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/( 
d*b)^(1/2))*a*b^3*c^3*d^3*x^2-42*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c 
^3*d^2*x^2+24*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b* 
c)/(d*b)^(1/2))*a^3*b*c^2*d^4*x+108*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1 
/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*c^3*d^3*x-360*ln(1/2*(2*b*d* 
x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*c^4*d^ 
2*x-280*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^4*d*x-30*(d*b)^(1/2)*((b 
*x+a)*(d*x+c))^(1/2)*a^2*b*c^3*d^2+290*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2) 
*a*b^2*c^4*d-48*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^2*d^3*x-18*((b 
*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^3*d^5*x^2+18*ln(1/2*(2*b*d*x+2*((b*x+a) 
*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^4*c*d^5*x-18*((b*x+a)* 
(d*x+c))^(1/2)*(d*b)^(1/2)*a^3*c^2*d^3+9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c 
))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^4*d^6*x^2+9*ln(1/2*(2*b*d*x+2 
*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^4*c^2*d^4-6*( 
d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c*d^4*x^2+66*(d*b)^(1/2)*((b*x...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 565 vs. \(2 (180) = 360\).

Time = 0.41 (sec) , antiderivative size = 1144, normalized size of antiderivative = 5.25 \[ \int \frac {x^4}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[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...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 514 vs. \(2 (180) = 360\).

Time = 0.20 (sec) , antiderivative size = 514, normalized size of antiderivative = 2.36 \[ \int \frac {x^4}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {\frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{11} c^{2} d^{6} - 2 \, a b^{10} c d^{7} + a^{2} b^{9} d^{8}\right )} {\left (b x + a\right )}}{b^{7} c^{2} d^{7} {\left | b \right |} - 2 \, a b^{6} c d^{8} {\left | b \right |} + a^{2} b^{5} d^{9} {\left | b \right |}} - \frac {7 \, b^{12} c^{3} d^{5} - 5 \, a b^{11} c^{2} d^{6} - 11 \, a^{2} b^{10} c d^{7} + 9 \, a^{3} b^{9} d^{8}}{b^{7} c^{2} d^{7} {\left | b \right |} - 2 \, a b^{6} c d^{8} {\left | b \right |} + a^{2} b^{5} d^{9} {\left | b \right |}}\right )} - \frac {4 \, {\left (35 \, b^{13} c^{4} d^{4} - 60 \, a b^{12} c^{3} d^{5} + 18 \, a^{2} b^{11} c^{2} d^{6} + 12 \, a^{3} b^{10} c d^{7} - 9 \, a^{4} b^{9} d^{8}\right )}}{b^{7} c^{2} d^{7} {\left | b \right |} - 2 \, a b^{6} c d^{8} {\left | b \right |} + a^{2} b^{5} d^{9} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (35 \, b^{14} c^{5} d^{3} - 95 \, a b^{13} c^{4} d^{4} + 78 \, a^{2} b^{12} c^{3} d^{5} - 14 \, a^{3} b^{11} c^{2} d^{6} - 9 \, a^{4} b^{10} c d^{7} + 5 \, a^{5} b^{9} d^{8}\right )}}{b^{7} c^{2} d^{7} {\left | b \right |} - 2 \, a b^{6} c d^{8} {\left | b \right |} + a^{2} b^{5} d^{9} {\left | b \right |}}\right )} \sqrt {b x + a}}{{\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {3 \, {\left (35 \, b^{5} c^{2} + 10 \, a b^{4} c d + 3 \, a^{2} b^{3} d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{4} {\left | b \right |}}}{12 \, b^{4}} \] Input:

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

Output:

1/12*(((3*(b*x + a)*(2*(b^11*c^2*d^6 - 2*a*b^10*c*d^7 + a^2*b^9*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^12*c^3*d^5 - 5*a*b^11*c^2*d^6 - 11*a^2*b^10*c*d^7 + 9*a^3*b^9*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^13 
*c^4*d^4 - 60*a*b^12*c^3*d^5 + 18*a^2*b^11*c^2*d^6 + 12*a^3*b^10*c*d^7 - 9 
*a^4*b^9*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^14*c^5*d^3 - 95*a*b^13*c^4*d^4 + 78*a^2*b^12*c^3 
*d^5 - 14*a^3*b^11*c^2*d^6 - 9*a^4*b^10*c*d^7 + 5*a^5*b^9*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) - 3*(35*b^5*c^2 + 10*a*b^4*c*d + 3*a^2*b^ 
3*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)*d^4*abs(b)))/b^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 13.39 (sec) , antiderivative size = 1449, normalized size of antiderivative = 6.65 \[ \int \frac {x^4}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 72*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*c**2*d**4 - 144*sqrt(c + d*x)*sq 
rt(a + b*x)*a**3*b*c*d**5*x - 72*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*d**6*x 
**2 - 120*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c**3*d**3 - 192*sqrt(c + d 
*x)*sqrt(a + b*x)*a**2*b**2*c**2*d**4*x - 24*sqrt(c + d*x)*sqrt(a + b*x)*a 
**2*b**2*c*d**5*x**2 + 48*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*d**6*x**3 
+ 1160*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c**4*d**2 + 1584*sqrt(c + d*x)*s 
qrt(a + b*x)*a*b**3*c**3*d**3*x + 264*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c 
**2*d**4*x**2 - 96*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c*d**5*x**3 - 840*sq 
rt(c + d*x)*sqrt(a + b*x)*b**4*c**5*d - 1120*sqrt(c + d*x)*sqrt(a + b*x)*b 
**4*c**4*d**2*x - 168*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**3*d**3*x**2 + 48 
*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**2*d**4*x**3 + 72*sqrt(d)*sqrt(b)*log( 
(sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*c**2 
*d**4 + 144*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + 
d*x))/sqrt(a*d - b*c))*a**4*c*d**5*x + 72*sqrt(d)*sqrt(b)*log((sqrt(d)*sqr 
t(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*d**6*x**2 + 96*s 
qrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a* 
d - b*c))*a**3*b*c**3*d**3 + 192*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x 
) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b*c**2*d**4*x + 96*sqrt(d 
)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b 
*c))*a**3*b*c*d**5*x**2 + 432*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x...