\(\int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx\) [86]

Optimal result

Integrand size = 36, antiderivative size = 219 \[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=-\frac {2 f \left (3 d^2 e^2+c^2 f^2\right ) \sqrt {b c^2-b d^2 x^2}}{b d^4 \sqrt {c+d x}}+\frac {2 f^2 (3 d e+c f) \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b^2 d^4 (c+d x)^{3/2}}-\frac {2 f^3 \left (b c^2-b d^2 x^2\right )^{5/2}}{5 b^3 d^4 (c+d x)^{5/2}}-\frac {\sqrt {2} (d e-c f)^3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}{\sqrt {b c^2-b d^2 x^2}}\right )}{\sqrt {b} \sqrt {c} d^4} \] Output:

-2*f*(c^2*f^2+3*d^2*e^2)*(-b*d^2*x^2+b*c^2)^(1/2)/b/d^4/(d*x+c)^(1/2)+2/3* 
f^2*(c*f+3*d*e)*(-b*d^2*x^2+b*c^2)^(3/2)/b^2/d^4/(d*x+c)^(3/2)-2/5*f^3*(-b 
*d^2*x^2+b*c^2)^(5/2)/b^3/d^4/(d*x+c)^(5/2)-2^(1/2)*(-c*f+d*e)^3*arctanh(2 
^(1/2)*b^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2))/b^(1/2)/c^( 
1/2)/d^4
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.79 \[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-\frac {2 f \sqrt {c^2-d^2 x^2} \left (13 c^2 f^2-c d f (15 e+f x)+3 d^2 \left (15 e^2+5 e f x+f^2 x^2\right )\right )}{\sqrt {c+d x}}+\frac {15 \sqrt {2} (-d e+c f)^3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}\right )}{15 d^4 \sqrt {b \left (c^2-d^2 x^2\right )}} \] Input:

Integrate[(e + f*x)^3/(Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2]),x]
 

Output:

(Sqrt[c^2 - d^2*x^2]*((-2*f*Sqrt[c^2 - d^2*x^2]*(13*c^2*f^2 - c*d*f*(15*e 
+ f*x) + 3*d^2*(15*e^2 + 5*e*f*x + f^2*x^2)))/Sqrt[c + d*x] + (15*Sqrt[2]* 
(-(d*e) + c*f)^3*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^ 
2]])/Sqrt[c]))/(15*d^4*Sqrt[b*(c^2 - d^2*x^2)])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {711, 27, 2170, 27, 600, 458, 471, 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.

\(\displaystyle \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx\)

\(\Big \downarrow \) 711

\(\displaystyle -\frac {2 \int -\frac {b f^2 (15 d e-7 c f) x^2 d^4+b f \left (15 d^2 e^2+c^2 f^2\right ) x d^3+b \left (5 d^3 e^3+3 c^3 f^3\right ) d^2}{2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b f^2 (15 d e-7 c f) x^2 d^4+b f \left (15 d^2 e^2+c^2 f^2\right ) x d^3+b \left (5 d^3 e^3+3 c^3 f^3\right ) d^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {b^2 d^6 \left (15 d^3 e^3+15 c^2 d f^2 e+2 c^3 f^3+d f \left (45 d^2 e^2-30 c d f e+17 c^2 f^2\right ) x\right )}{2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx}{3 b d^4}-\frac {2}{3} d f^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} (15 d e-7 c f)}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} b d^2 \int \frac {15 d^3 e^3+15 c^2 d f^2 e+2 c^3 f^3+d f \left (45 d^2 e^2-30 c d f e+17 c^2 f^2\right ) x}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx-\frac {2}{3} d f^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} (15 d e-7 c f)}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 600

\(\displaystyle \frac {\frac {1}{3} b d^2 \left (f \left (17 c^2 f^2-30 c d e f+45 d^2 e^2\right ) \int \frac {\sqrt {c+d x}}{\sqrt {b c^2-b d^2 x^2}}dx+15 (d e-c f)^3 \int \frac {1}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx\right )-\frac {2}{3} d f^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} (15 d e-7 c f)}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {1}{3} b d^2 \left (15 (d e-c f)^3 \int \frac {1}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx-\frac {2 f \sqrt {b c^2-b d^2 x^2} \left (17 c^2 f^2-30 c d e f+45 d^2 e^2\right )}{b d \sqrt {c+d x}}\right )-\frac {2}{3} d f^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} (15 d e-7 c f)}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {\frac {1}{3} b d^2 \left (30 d (d e-c f)^3 \int \frac {1}{\frac {d^2 \left (b c^2-b d^2 x^2\right )}{c+d x}-2 b c d^2}d\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}-\frac {2 f \sqrt {b c^2-b d^2 x^2} \left (17 c^2 f^2-30 c d e f+45 d^2 e^2\right )}{b d \sqrt {c+d x}}\right )-\frac {2}{3} d f^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} (15 d e-7 c f)}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{3} b d^2 \left (-\frac {15 \sqrt {2} (d e-c f)^3 \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d}-\frac {2 f \sqrt {b c^2-b d^2 x^2} \left (17 c^2 f^2-30 c d e f+45 d^2 e^2\right )}{b d \sqrt {c+d x}}\right )-\frac {2}{3} d f^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} (15 d e-7 c f)}{5 b d^5}-\frac {2 f^3 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{5 b d^4}\)

Input:

Int[(e + f*x)^3/(Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2]),x]
 

Output:

(-2*f^3*(c + d*x)^(3/2)*Sqrt[b*c^2 - b*d^2*x^2])/(5*b*d^4) + ((-2*d*f^2*(1 
5*d*e - 7*c*f)*Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2])/3 + (b*d^2*((-2*f*(4 
5*d^2*e^2 - 30*c*d*e*f + 17*c^2*f^2)*Sqrt[b*c^2 - b*d^2*x^2])/(b*d*Sqrt[c 
+ d*x]) - (15*Sqrt[2]*(d*e - c*f)^3*ArcTanh[Sqrt[b*c^2 - b*d^2*x^2]/(Sqrt[ 
2]*Sqrt[b]*Sqrt[c]*Sqrt[c + d*x])])/(Sqrt[b]*Sqrt[c]*d)))/3)/(5*b*d^5)
 

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 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]
 

rule 458

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

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 600

Int[((A_.) + (B_.)*(x_))/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2] 
), x_Symbol] :> Simp[B/d   Int[Sqrt[c + d*x]/Sqrt[a + b*x^2], x], x] - Simp 
[(B*c - A*d)/d   Int[1/(Sqrt[c + d*x]*Sqrt[a + b*x^2]), x], x] /; FreeQ[{a, 
 b, c, d, A, B}, x] && NegQ[b/a]
 

rule 711

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (c_.)*(x_ 
)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + c*x^2)^(p + 1) 
/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) 
Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x) 
^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - 2*e*g^n*(m + p + n)*(d + e*x)^(n 
 - 2)*(a*e - c*d*x), x], x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && Eq 
Q[c*d^2 + a*e^2, 0] && IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
 

rule 2170

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

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {2 f \left (3 d^{2} f^{2} x^{2}-c d \,f^{2} x +15 d^{2} e f x +13 c^{2} f^{2}-15 c d e f +45 d^{2} e^{2}\right ) \left (-d x +c \right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}}{15 d^{4} \sqrt {-b \left (d x -c \right )}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}+\frac {\left (f^{3} c^{3}-3 c^{2} d e \,f^{2}+3 e^{2} f \,d^{2} c -e^{3} d^{3}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-b d x +b c}\, \sqrt {2}}{2 \sqrt {b c}}\right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}}{d^{4} \sqrt {b c}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(246\)
default	\(\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) b \,c^{3} f^{3}-45 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) b \,c^{2} d e \,f^{2}+45 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) b c \,d^{2} e^{2} f -15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) b \,d^{3} e^{3}-6 d^{2} f^{3} x^{2} \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}+2 c d \,f^{3} x \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-30 d^{2} e \,f^{2} x \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-26 c^{2} f^{3} \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}+30 c d e \,f^{2} \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-90 d^{2} e^{2} f \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}\right )}{15 b \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b}\, d^{4} \sqrt {b c}}\)	\(328\)

Input:

int((f*x+e)^3/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2),x,method=_RETURNVERBO 
SE)
 

Output:

-2/15*f*(3*d^2*f^2*x^2-c*d*f^2*x+15*d^2*e*f*x+13*c^2*f^2-15*c*d*e*f+45*d^2 
*e^2)*(-d*x+c)/d^4/(-b*(d*x-c))^(1/2)*(-1/(d*x+c)*b*(d^2*x^2-c^2))^(1/2)*( 
d*x+c)^(1/2)/(-b*(d^2*x^2-c^2))^(1/2)+(c^3*f^3-3*c^2*d*e*f^2+3*c*d^2*e^2*f 
-d^3*e^3)/d^4*2^(1/2)/(b*c)^(1/2)*arctanh(1/2*(-b*d*x+b*c)^(1/2)*2^(1/2)/( 
b*c)^(1/2))*(-1/(d*x+c)*b*(d^2*x^2-c^2))^(1/2)*(d*x+c)^(1/2)/(-b*(d^2*x^2- 
c^2))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.46 \[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (b c d^{3} e^{3} - 3 \, b c^{2} d^{2} e^{2} f + 3 \, b c^{3} d e f^{2} - b c^{4} f^{3} + {\left (b d^{4} e^{3} - 3 \, b c d^{3} e^{2} f + 3 \, b c^{2} d^{2} e f^{2} - b c^{3} d f^{3}\right )} x\right )} \sqrt {\frac {1}{b c}} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} c \sqrt {\frac {1}{b c}} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (3 \, d^{2} f^{3} x^{2} + 45 \, d^{2} e^{2} f - 15 \, c d e f^{2} + 13 \, c^{2} f^{3} + {\left (15 \, d^{2} e f^{2} - c d f^{3}\right )} x\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{30 \, {\left (b d^{5} x + b c d^{4}\right )}}, -\frac {15 \, \sqrt {2} {\left (b c d^{3} e^{3} - 3 \, b c^{2} d^{2} e^{2} f + 3 \, b c^{3} d e f^{2} - b c^{4} f^{3} + {\left (b d^{4} e^{3} - 3 \, b c d^{3} e^{2} f + 3 \, b c^{2} d^{2} e f^{2} - b c^{3} d f^{3}\right )} x\right )} \sqrt {-\frac {1}{b c}} \arctan \left (\frac {\sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} c \sqrt {-\frac {1}{b c}}}{d^{2} x^{2} - c^{2}}\right ) + 2 \, {\left (3 \, d^{2} f^{3} x^{2} + 45 \, d^{2} e^{2} f - 15 \, c d e f^{2} + 13 \, c^{2} f^{3} + {\left (15 \, d^{2} e f^{2} - c d f^{3}\right )} x\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{15 \, {\left (b d^{5} x + b c d^{4}\right )}}\right ] \] Input:

integrate((f*x+e)^3/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="f 
ricas")
 

Output:

[-1/30*(15*sqrt(2)*(b*c*d^3*e^3 - 3*b*c^2*d^2*e^2*f + 3*b*c^3*d*e*f^2 - b* 
c^4*f^3 + (b*d^4*e^3 - 3*b*c*d^3*e^2*f + 3*b*c^2*d^2*e*f^2 - b*c^3*d*f^3)* 
x)*sqrt(1/(b*c))*log(-(d^2*x^2 - 2*c*d*x - 2*sqrt(2)*sqrt(-b*d^2*x^2 + b*c 
^2)*sqrt(d*x + c)*c*sqrt(1/(b*c)) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) + 4* 
(3*d^2*f^3*x^2 + 45*d^2*e^2*f - 15*c*d*e*f^2 + 13*c^2*f^3 + (15*d^2*e*f^2 
- c*d*f^3)*x)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c))/(b*d^5*x + b*c*d^4), 
 -1/15*(15*sqrt(2)*(b*c*d^3*e^3 - 3*b*c^2*d^2*e^2*f + 3*b*c^3*d*e*f^2 - b* 
c^4*f^3 + (b*d^4*e^3 - 3*b*c*d^3*e^2*f + 3*b*c^2*d^2*e*f^2 - b*c^3*d*f^3)* 
x)*sqrt(-1/(b*c))*arctan(sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*c* 
sqrt(-1/(b*c))/(d^2*x^2 - c^2)) + 2*(3*d^2*f^3*x^2 + 45*d^2*e^2*f - 15*c*d 
*e*f^2 + 13*c^2*f^3 + (15*d^2*e*f^2 - c*d*f^3)*x)*sqrt(-b*d^2*x^2 + b*c^2) 
*sqrt(d*x + c))/(b*d^5*x + b*c*d^4)]
 

Sympy [F]

\[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\int \frac {\left (e + f x\right )^{3}}{\sqrt {- b \left (- c + d x\right ) \left (c + d x\right )} \sqrt {c + d x}}\, dx \] Input:

integrate((f*x+e)**3/(d*x+c)**(1/2)/(-b*d**2*x**2+b*c**2)**(1/2),x)
 

Output:

Integral((e + f*x)**3/(sqrt(-b*(-c + d*x)*(c + d*x))*sqrt(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\int { \frac {{\left (f x + e\right )}^{3}}{\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}} \,d x } \] Input:

integrate((f*x+e)^3/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="m 
axima")
 

Output:

integrate((f*x + e)^3/(sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.03 \[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{2 \, \sqrt {-b c}}\right )}{\sqrt {-b c}} - \frac {2 \, {\left (45 \, \sqrt {-{\left (d x + c\right )} b + 2 \, b c} b^{14} d^{2} e^{2} f + 15 \, \sqrt {-{\left (d x + c\right )} b + 2 \, b c} b^{14} c^{2} f^{3} - 15 \, {\left (-{\left (d x + c\right )} b + 2 \, b c\right )}^{\frac {3}{2}} b^{13} d e f^{2} - 5 \, {\left (-{\left (d x + c\right )} b + 2 \, b c\right )}^{\frac {3}{2}} b^{13} c f^{3} + 3 \, {\left ({\left (d x + c\right )} b - 2 \, b c\right )}^{2} \sqrt {-{\left (d x + c\right )} b + 2 \, b c} b^{12} f^{3}\right )}}{b^{15}}}{15 \, d^{4}} \] Input:

integrate((f*x+e)^3/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="g 
iac")
 

Output:

1/15*(15*sqrt(2)*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*arcta 
n(1/2*sqrt(2)*sqrt(-(d*x + c)*b + 2*b*c)/sqrt(-b*c))/sqrt(-b*c) - 2*(45*sq 
rt(-(d*x + c)*b + 2*b*c)*b^14*d^2*e^2*f + 15*sqrt(-(d*x + c)*b + 2*b*c)*b^ 
14*c^2*f^3 - 15*(-(d*x + c)*b + 2*b*c)^(3/2)*b^13*d*e*f^2 - 5*(-(d*x + c)* 
b + 2*b*c)^(3/2)*b^13*c*f^3 + 3*((d*x + c)*b - 2*b*c)^2*sqrt(-(d*x + c)*b 
+ 2*b*c)*b^12*f^3)/b^15)/d^4
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\int \frac {{\left (e+f\,x\right )}^3}{\sqrt {b\,c^2-b\,d^2\,x^2}\,\sqrt {c+d\,x}} \,d x \] Input:

int((e + f*x)^3/((b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2)),x)
 

Output:

int((e + f*x)^3/((b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.54 \[ \int \frac {(e+f x)^3}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\sqrt {b}\, \left (-52 \sqrt {-d x +c}\, c^{3} f^{3}+60 \sqrt {-d x +c}\, c^{2} d e \,f^{2}+4 \sqrt {-d x +c}\, c^{2} d \,f^{3} x -180 \sqrt {-d x +c}\, c \,d^{2} e^{2} f -60 \sqrt {-d x +c}\, c \,d^{2} e \,f^{2} x -12 \sqrt {-d x +c}\, c \,d^{2} f^{3} x^{2}-15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{3} f^{3}+45 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{2} d e \,f^{2}-45 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c \,d^{2} e^{2} f +15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) d^{3} e^{3}+15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{3} f^{3}-45 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{2} d e \,f^{2}+45 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c \,d^{2} e^{2} f -15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) d^{3} e^{3}\right )}{30 b c \,d^{4}} \] Input:

int((f*x+e)^3/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2),x)
 

Output:

(sqrt(b)*( - 52*sqrt(c - d*x)*c**3*f**3 + 60*sqrt(c - d*x)*c**2*d*e*f**2 + 
 4*sqrt(c - d*x)*c**2*d*f**3*x - 180*sqrt(c - d*x)*c*d**2*e**2*f - 60*sqrt 
(c - d*x)*c*d**2*e*f**2*x - 12*sqrt(c - d*x)*c*d**2*f**3*x**2 - 15*sqrt(c) 
*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**3*f**3 + 45*sqrt(c)*sqrt( 
2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**2*d*e*f**2 - 45*sqrt(c)*sqrt(2) 
*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c*d**2*e**2*f + 15*sqrt(c)*sqrt(2)*l 
og(sqrt(c - d*x) - sqrt(c)*sqrt(2))*d**3*e**3 + 15*sqrt(c)*sqrt(2)*log(sqr 
t(c - d*x) + sqrt(c)*sqrt(2))*c**3*f**3 - 45*sqrt(c)*sqrt(2)*log(sqrt(c - 
d*x) + sqrt(c)*sqrt(2))*c**2*d*e*f**2 + 45*sqrt(c)*sqrt(2)*log(sqrt(c - d* 
x) + sqrt(c)*sqrt(2))*c*d**2*e**2*f - 15*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) 
 + sqrt(c)*sqrt(2))*d**3*e**3))/(30*b*c*d**4)