\(\int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx\) [1181]

Optimal result

Integrand size = 20, antiderivative size = 204 \[ \int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx=\frac {\left (5 b^2 c^4-30 a b c^2 d^2+4 a^2 d^4\right ) \sqrt {a+b x^2}}{5 b^3}+\frac {c d \left (4 b c^2-3 a d^2\right ) x \sqrt {a+b x^2}}{2 b^2}+\frac {c d^3 x^3 \sqrt {a+b x^2}}{b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}+\frac {2 d^2 \left (15 b c^2-2 a d^2\right ) \left (a+b x^2\right )^{3/2}}{15 b^3}-\frac {a c d \left (4 b c^2-3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:

1/5*(4*a^2*d^4-30*a*b*c^2*d^2+5*b^2*c^4)*(b*x^2+a)^(1/2)/b^3+1/2*c*d*(-3*a 
*d^2+4*b*c^2)*x*(b*x^2+a)^(1/2)/b^2+c*d^3*x^3*(b*x^2+a)^(1/2)/b+1/5*d^4*x^ 
4*(b*x^2+a)^(1/2)/b+2/15*d^2*(-2*a*d^2+15*b*c^2)*(b*x^2+a)^(3/2)/b^3-1/2*a 
*c*d*(-3*a*d^2+4*b*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.71 \[ \int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (16 a^2 d^4-a b d^2 \left (120 c^2+45 c d x+8 d^2 x^2\right )+6 b^2 \left (5 c^4+10 c^3 d x+10 c^2 d^2 x^2+5 c d^3 x^3+d^4 x^4\right )\right )+15 a \sqrt {b} c d \left (4 b c^2-3 a d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{30 b^3} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(16*a^2*d^4 - a*b*d^2*(120*c^2 + 45*c*d*x + 8*d^2*x^2) + 
6*b^2*(5*c^4 + 10*c^3*d*x + 10*c^2*d^2*x^2 + 5*c*d^3*x^3 + d^4*x^4)) + 15* 
a*Sqrt[b]*c*d*(4*b*c^2 - 3*a*d^2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(30 
*b^3)
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {541, 2340, 27, 2340, 27, 533, 27, 455, 224, 219}

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.

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

\(\Big \downarrow \) 541

\(\displaystyle \frac {\int \frac {x \left (5 b c^4+20 b d x c^3+20 b d^3 x^3 c+2 d^2 \left (15 b c^2-2 a d^2\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 2340

\(\displaystyle \frac {\frac {\int \frac {4 x \left (5 b^2 c^4+5 b d \left (4 b c^2-3 a d^2\right ) x c+2 b d^2 \left (15 b c^2-2 a d^2\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{4 b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {x \left (5 b^2 c^4+5 b d \left (4 b c^2-3 a d^2\right ) x c+2 b d^2 \left (15 b c^2-2 a d^2\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 2340

\(\displaystyle \frac {\frac {\frac {\int \frac {b x \left (15 b^2 c^4-60 a b d^2 c^2+15 b d \left (4 b c^2-3 a d^2\right ) x c+8 a^2 d^4\right )}{\sqrt {b x^2+a}}dx}{3 b}+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{3} \int \frac {x \left (15 b^2 c^4-60 a b d^2 c^2+15 b d \left (4 b c^2-3 a d^2\right ) x c+8 a^2 d^4\right )}{\sqrt {b x^2+a}}dx+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {15}{2} c d x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )-\frac {\int \frac {b \left (15 a c d \left (4 b c^2-3 a d^2\right )-2 \left (15 b^2 c^4-60 a b d^2 c^2+8 a^2 d^4\right ) x\right )}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {15}{2} c d x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )-\frac {1}{2} \int \frac {15 a c d \left (4 b c^2-3 a d^2\right )-2 \left (15 b^2 c^4-60 a b d^2 c^2+8 a^2 d^4\right ) x}{\sqrt {b x^2+a}}dx\right )+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 \sqrt {a+b x^2} \left (8 a^2 d^4-60 a b c^2 d^2+15 b^2 c^4\right )}{b}-15 a c d \left (4 b c^2-3 a d^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx\right )+\frac {15}{2} c d x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )\right )+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 \sqrt {a+b x^2} \left (8 a^2 d^4-60 a b c^2 d^2+15 b^2 c^4\right )}{b}-15 a c d \left (4 b c^2-3 a d^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )+\frac {15}{2} c d x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )\right )+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 \sqrt {a+b x^2} \left (8 a^2 d^4-60 a b c^2 d^2+15 b^2 c^4\right )}{b}-\frac {15 a c d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (4 b c^2-3 a d^2\right )}{\sqrt {b}}\right )+\frac {15}{2} c d x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )\right )+\frac {2}{3} d^2 x^2 \sqrt {a+b x^2} \left (15 b c^2-2 a d^2\right )}{b}+5 c d^3 x^3 \sqrt {a+b x^2}}{5 b}+\frac {d^4 x^4 \sqrt {a+b x^2}}{5 b}\)

Input:

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

Output:

(d^4*x^4*Sqrt[a + b*x^2])/(5*b) + (5*c*d^3*x^3*Sqrt[a + b*x^2] + ((2*d^2*( 
15*b*c^2 - 2*a*d^2)*x^2*Sqrt[a + b*x^2])/3 + ((15*c*d*(4*b*c^2 - 3*a*d^2)* 
x*Sqrt[a + b*x^2])/2 + ((2*(15*b^2*c^4 - 60*a*b*c^2*d^2 + 8*a^2*d^4)*Sqrt[ 
a + b*x^2])/b - (15*a*c*d*(4*b*c^2 - 3*a*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + 
 b*x^2]])/Sqrt[b])/2)/3)/b)/(5*b)
 

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 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {\left (6 b^{2} d^{4} x^{4}+30 b^{2} c \,d^{3} x^{3}-8 a b \,d^{4} x^{2}+60 d^{2} c^{2} x^{2} b^{2}-45 a b c \,d^{3} x +60 b^{2} c^{3} d x +16 a^{2} d^{4}-120 b \,c^{2} d^{2} a +30 b^{2} c^{4}\right ) \sqrt {b \,x^{2}+a}}{30 b^{3}}+\frac {a c d \left (3 a \,d^{2}-4 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {5}{2}}}\)	\(146\)
default	\(\frac {c^{4} \sqrt {b \,x^{2}+a}}{b}+d^{4} \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )+4 c \,d^{3} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+4 d \,c^{3} \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+6 c^{2} d^{2} \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )\)	\(232\)

Input:

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

Output:

1/30*(6*b^2*d^4*x^4+30*b^2*c*d^3*x^3-8*a*b*d^4*x^2+60*b^2*c^2*d^2*x^2-45*a 
*b*c*d^3*x+60*b^2*c^3*d*x+16*a^2*d^4-120*a*b*c^2*d^2+30*b^2*c^4)*(b*x^2+a) 
^(1/2)/b^3+1/2*a/b^(5/2)*c*d*(3*a*d^2-4*b*c^2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2 
))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.54 \[ \int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx=\left [-\frac {15 \, {\left (4 \, a b c^{3} d - 3 \, a^{2} c d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (6 \, b^{2} d^{4} x^{4} + 30 \, b^{2} c d^{3} x^{3} + 30 \, b^{2} c^{4} - 120 \, a b c^{2} d^{2} + 16 \, a^{2} d^{4} + 4 \, {\left (15 \, b^{2} c^{2} d^{2} - 2 \, a b d^{4}\right )} x^{2} + 15 \, {\left (4 \, b^{2} c^{3} d - 3 \, a b c d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{60 \, b^{3}}, \frac {15 \, {\left (4 \, a b c^{3} d - 3 \, a^{2} c d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, b^{2} d^{4} x^{4} + 30 \, b^{2} c d^{3} x^{3} + 30 \, b^{2} c^{4} - 120 \, a b c^{2} d^{2} + 16 \, a^{2} d^{4} + 4 \, {\left (15 \, b^{2} c^{2} d^{2} - 2 \, a b d^{4}\right )} x^{2} + 15 \, {\left (4 \, b^{2} c^{3} d - 3 \, a b c d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{30 \, b^{3}}\right ] \] Input:

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

Output:

[-1/60*(15*(4*a*b*c^3*d - 3*a^2*c*d^3)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 
 + a)*sqrt(b)*x - a) - 2*(6*b^2*d^4*x^4 + 30*b^2*c*d^3*x^3 + 30*b^2*c^4 - 
120*a*b*c^2*d^2 + 16*a^2*d^4 + 4*(15*b^2*c^2*d^2 - 2*a*b*d^4)*x^2 + 15*(4* 
b^2*c^3*d - 3*a*b*c*d^3)*x)*sqrt(b*x^2 + a))/b^3, 1/30*(15*(4*a*b*c^3*d - 
3*a^2*c*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (6*b^2*d^4*x^4 
+ 30*b^2*c*d^3*x^3 + 30*b^2*c^4 - 120*a*b*c^2*d^2 + 16*a^2*d^4 + 4*(15*b^2 
*c^2*d^2 - 2*a*b*d^4)*x^2 + 15*(4*b^2*c^3*d - 3*a*b*c*d^3)*x)*sqrt(b*x^2 + 
 a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.13 \[ \int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (- \frac {3 a c d^{3}}{b} + 4 c^{3} d\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {c d^{3} x^{3}}{b} + \frac {d^{4} x^{4}}{5 b} + \frac {x^{2} \left (- \frac {4 a d^{4}}{5 b} + 6 c^{2} d^{2}\right )}{3 b} + \frac {x \left (- \frac {3 a c d^{3}}{b} + 4 c^{3} d\right )}{2 b} + \frac {- \frac {2 a \left (- \frac {4 a d^{4}}{5 b} + 6 c^{2} d^{2}\right )}{3 b} + c^{4}}{b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {c^{4} x^{2}}{2} + \frac {4 c^{3} d x^{3}}{3} + \frac {3 c^{2} d^{2} x^{4}}{2} + \frac {4 c d^{3} x^{5}}{5} + \frac {d^{4} x^{6}}{6}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a*(-3*a*c*d**3/b + 4*c**3*d)*Piecewise((log(2*sqrt(b)*sqrt(a + 
 b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(2*b) 
 + sqrt(a + b*x**2)*(c*d**3*x**3/b + d**4*x**4/(5*b) + x**2*(-4*a*d**4/(5* 
b) + 6*c**2*d**2)/(3*b) + x*(-3*a*c*d**3/b + 4*c**3*d)/(2*b) + (-2*a*(-4*a 
*d**4/(5*b) + 6*c**2*d**2)/(3*b) + c**4)/b), Ne(b, 0)), ((c**4*x**2/2 + 4* 
c**3*d*x**3/3 + 3*c**2*d**2*x**4/2 + 4*c*d**3*x**5/5 + d**4*x**6/6)/sqrt(a 
), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.08 \[ \int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d^{4} x^{4}}{5 \, b} + \frac {\sqrt {b x^{2} + a} c d^{3} x^{3}}{b} + \frac {2 \, \sqrt {b x^{2} + a} c^{2} d^{2} x^{2}}{b} - \frac {4 \, \sqrt {b x^{2} + a} a d^{4} x^{2}}{15 \, b^{2}} + \frac {2 \, \sqrt {b x^{2} + a} c^{3} d x}{b} - \frac {3 \, \sqrt {b x^{2} + a} a c d^{3} x}{2 \, b^{2}} - \frac {2 \, a c^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, a^{2} c d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a} c^{4}}{b} - \frac {4 \, \sqrt {b x^{2} + a} a c^{2} d^{2}}{b^{2}} + \frac {8 \, \sqrt {b x^{2} + a} a^{2} d^{4}}{15 \, b^{3}} \] Input:

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

Output:

1/5*sqrt(b*x^2 + a)*d^4*x^4/b + sqrt(b*x^2 + a)*c*d^3*x^3/b + 2*sqrt(b*x^2 
 + a)*c^2*d^2*x^2/b - 4/15*sqrt(b*x^2 + a)*a*d^4*x^2/b^2 + 2*sqrt(b*x^2 + 
a)*c^3*d*x/b - 3/2*sqrt(b*x^2 + a)*a*c*d^3*x/b^2 - 2*a*c^3*d*arcsinh(b*x/s 
qrt(a*b))/b^(3/2) + 3/2*a^2*c*d^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) + sqrt(b* 
x^2 + a)*c^4/b - 4*sqrt(b*x^2 + a)*a*c^2*d^2/b^2 + 8/15*sqrt(b*x^2 + a)*a^ 
2*d^4/b^3
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83 \[ \int \frac {x (c+d x)^4}{\sqrt {a+b x^2}} \, dx=\frac {1}{30} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (\frac {d^{4} x}{b} + \frac {5 \, c d^{3}}{b}\right )} x + \frac {2 \, {\left (15 \, b^{4} c^{2} d^{2} - 2 \, a b^{3} d^{4}\right )}}{b^{5}}\right )} x + \frac {15 \, {\left (4 \, b^{4} c^{3} d - 3 \, a b^{3} c d^{3}\right )}}{b^{5}}\right )} x + \frac {2 \, {\left (15 \, b^{4} c^{4} - 60 \, a b^{3} c^{2} d^{2} + 8 \, a^{2} b^{2} d^{4}\right )}}{b^{5}}\right )} + \frac {{\left (4 \, a b c^{3} d - 3 \, a^{2} c d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/30*sqrt(b*x^2 + a)*((2*(3*(d^4*x/b + 5*c*d^3/b)*x + 2*(15*b^4*c^2*d^2 - 
2*a*b^3*d^4)/b^5)*x + 15*(4*b^4*c^3*d - 3*a*b^3*c*d^3)/b^5)*x + 2*(15*b^4* 
c^4 - 60*a*b^3*c^2*d^2 + 8*a^2*b^2*d^4)/b^5) + 1/2*(4*a*b*c^3*d - 3*a^2*c* 
d^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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