\(\int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx\) [760]

Optimal result

Integrand size = 24, antiderivative size = 239 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt [4]{\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{\left (2 c d-b e+\sqrt {b^2-4 a c} e\right ) \sqrt {d+e x} \sqrt [4]{a+b x+c x^2}} \] Output:

2*(b-(-4*a*c+b^2)^(1/2)+2*c*x)*((2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)*(2*c*x+(- 
4*a*c+b^2)^(1/2)+b)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e)/(b-(-4*a*c+b^2)^(1/2) 
+2*c*x))^(1/4)*hypergeom([-1/2, 1/4],[1/2],-4*c*(-4*a*c+b^2)^(1/2)*(e*x+d) 
/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e)/(b-(-4*a*c+b^2)^(1/2)+2*c*x))/(2*c*d-b*e 
+(-4*a*c+b^2)^(1/2)*e)/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.36 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=-\frac {2 \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{3/4} \sqrt {d+e x} \sqrt [4]{a+x (b+c x)}} \] Input:

Integrate[1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)),x]
 

Output:

(-2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)*Hypergeometric2F1[-1/2, 1/4, 1/2, (-4* 
c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + 
 Sqrt[b^2 - 4*a*c] - 2*c*x))])/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(((2* 
c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((-2*c* 
d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)))^(3/4)*Sq 
rt[d + e*x]*(a + x*(b + c*x))^(1/4))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1155}

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.

\(\displaystyle \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 1155

\(\displaystyle \frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \sqrt [4]{\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {d+e x} \sqrt [4]{a+b x+c x^2} \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}\)

Input:

Int[1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)),x]
 

Output:

(2*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*( 
b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - 
Sqrt[b^2 - 4*a*c] + 2*c*x)))^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, (-4*c 
*Sqrt[b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sq 
rt[b^2 - 4*a*c] + 2*c*x))])/((2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)*Sqrt[d + 
e*x]*(a + b*x + c*x^2)^(1/4))
 

Defintions of rubi rules used

rule 1155

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(b - q + 2*c*x))*(d + e*x)^ 
(m + 1)*((a + b*x + c*x^2)^p/((m + 1)*(2*c*d - b*e + e*q)*((2*c*d - b*e + e 
*q)*((b + q + 2*c*x)/((2*c*d - b*e - e*q)*(b - q + 2*c*x))))^p))*Hypergeome 
tric2F1[m + 1, -p, m + 2, -4*c*q*((d + e*x)/((2*c*d - b*e - e*q)*(b - q + 2 
*c*x)))], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[m + 2*p + 2, 0]
 

Maple [F]

\[\int \frac {1}{\left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x\]

Input:

int(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x)
 

Output:

int(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x, algorithm="fricas")
 

Output:

integral((c*x^2 + b*x + a)^(3/4)*sqrt(e*x + d)/(c*e^2*x^4 + (2*c*d*e + b*e 
^2)*x^3 + a*d^2 + (c*d^2 + 2*b*d*e + a*e^2)*x^2 + (b*d^2 + 2*a*d*e)*x), x)
 

Sympy [F]

\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \sqrt [4]{a + b x + c x^{2}}}\, dx \] Input:

integrate(1/(e*x+d)**(3/2)/(c*x**2+b*x+a)**(1/4),x)
 

Output:

Integral(1/((d + e*x)**(3/2)*(a + b*x + c*x**2)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate(1/((c*x^2 + b*x + a)^(1/4)*(e*x + d)^(3/2)), x)
 

Giac [F]

\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x, algorithm="giac")
 

Output:

integrate(1/((c*x^2 + b*x + a)^(1/4)*(e*x + d)^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {1}{{\left (d+e\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^{1/4}} \,d x \] Input:

int(1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)),x)
 

Output:

int(1/((d + e*x)^(3/2)*(a + b*x + c*x^2)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {1}{\left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x \] Input:

int(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x)
 

Output:

int(1/(e*x+d)^(3/2)/(c*x^2+b*x+a)^(1/4),x)