3.422 \(\int \frac {(d+e x)^{7/2}}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=383 \[ \frac {2 d \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (c d-b e) \left (-b^2 e^2-16 b c d e+16 c^2 d^2\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{3 (-b)^{7/2} c^{3/2} \sqrt {b x+c x^2} \sqrt {d+e x}}-\frac {4 \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{3 (-b)^{7/2} c^{3/2} \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}-\frac {2 (d+e x)^{5/2} (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (x (2 c d-b e) \left (-b^2 e^2-8 b c d e+8 c^2 d^2\right )+b c d^2 (8 c d-9 b e)\right )}{3 b^4 c \sqrt {b x+c x^2}} \]

[Out]

-2/3*(e*x+d)^(5/2)*(b*d+(-b*e+2*c*d)*x)/b^2/(c*x^2+b*x)^(3/2)+2/3*(b*c*d^2*(-9*b*e+8*c*d)+(-b*e+2*c*d)*(-b^2*e
^2-8*b*c*d*e+8*c^2*d^2)*x)*(e*x+d)^(1/2)/b^4/c/(c*x^2+b*x)^(1/2)-4/3*(-b*e+2*c*d)*(-b^2*e^2-4*b*c*d*e+4*c^2*d^
2)*EllipticE(c^(1/2)*x^(1/2)/(-b)^(1/2),(b*e/c/d)^(1/2))*x^(1/2)*(c*x/b+1)^(1/2)*(e*x+d)^(1/2)/(-b)^(7/2)/c^(3
/2)/(1+e*x/d)^(1/2)/(c*x^2+b*x)^(1/2)+2/3*d*(-b*e+c*d)*(-b^2*e^2-16*b*c*d*e+16*c^2*d^2)*EllipticF(c^(1/2)*x^(1
/2)/(-b)^(1/2),(b*e/c/d)^(1/2))*x^(1/2)*(c*x/b+1)^(1/2)*(1+e*x/d)^(1/2)/(-b)^(7/2)/c^(3/2)/(e*x+d)^(1/2)/(c*x^
2+b*x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {738, 818, 843, 715, 112, 110, 117, 116} \[ \frac {2 \sqrt {d+e x} \left (x (2 c d-b e) \left (-b^2 e^2-8 b c d e+8 c^2 d^2\right )+b c d^2 (8 c d-9 b e)\right )}{3 b^4 c \sqrt {b x+c x^2}}+\frac {2 d \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (c d-b e) \left (-b^2 e^2-16 b c d e+16 c^2 d^2\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{3 (-b)^{7/2} c^{3/2} \sqrt {b x+c x^2} \sqrt {d+e x}}-\frac {4 \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{3 (-b)^{7/2} c^{3/2} \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}-\frac {2 (d+e x)^{5/2} (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(7/2)/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^(5/2)*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (2*Sqrt[d + e*x]*(b*c*d^2*(8*c*d -
9*b*e) + (2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e - b^2*e^2)*x))/(3*b^4*c*Sqrt[b*x + c*x^2]) - (4*(2*c*d - b*e)*(4
*c^2*d^2 - 4*b*c*d*e - b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqr
t[-b]], (b*e)/(c*d)])/(3*(-b)^(7/2)*c^(3/2)*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (2*d*(c*d - b*e)*(16*c^2*d^
2 - 16*b*c*d*e - b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[
-b]], (b*e)/(c*d)])/(3*(-b)^(7/2)*c^(3/2)*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

Rule 110

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Sqrt[e]*Rt[-(b/d)
, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/b, x] /; FreeQ[{b, c, d, e, f}, x] &&
NeQ[d*e - c*f, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !LtQ[-(b/d), 0]

Rule 112

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*x]*Sqrt[
1 + (d*x)/c])/(Sqrt[c + d*x]*Sqrt[1 + (f*x)/e]), Int[Sqrt[1 + (f*x)/e]/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]), x], x] /
; FreeQ[{b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
 && GtQ[c, 0] && GtQ[e, 0] && (PosQ[-(b/d)] || NegQ[-(b/f)])

Rule 117

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + (d*x)/c]
*Sqrt[1 + (f*x)/e])/(Sqrt[c + d*x]*Sqrt[e + f*x]), Int[1/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]*Sqrt[1 + (f*x)/e]), x],
x] /; FreeQ[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 715

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(Sqrt[x]*Sqrt[b + c*x])/Sqrt[
b*x + c*x^2], Int[(d + e*x)^m/(Sqrt[x]*Sqrt[b + c*x]), x], x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] &
& NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
 4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
 e, m, p, x]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{7/2}}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^{5/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {(d+e x)^{3/2} \left (\frac {1}{2} d (8 c d-9 b e)-\frac {1}{2} e (2 c d-b e) x\right )}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac {2 (d+e x)^{5/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (b c d^2 (8 c d-9 b e)+(2 c d-b e) \left (8 c^2 d^2-8 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {4 \int \frac {\frac {1}{4} b d e \left (8 c^2 d^2-11 b c d e+b^2 e^2\right )+\frac {1}{2} e (2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x}{\sqrt {d+e x} \sqrt {b x+c x^2}} \, dx}{3 b^4 c}\\ &=-\frac {2 (d+e x)^{5/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (b c d^2 (8 c d-9 b e)+(2 c d-b e) \left (8 c^2 d^2-8 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}+\frac {\left (d (c d-b e) \left (16 c^2 d^2-16 b c d e-b^2 e^2\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {b x+c x^2}} \, dx}{3 b^4 c}-\frac {\left (2 (2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {b x+c x^2}} \, dx}{3 b^4 c}\\ &=-\frac {2 (d+e x)^{5/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (b c d^2 (8 c d-9 b e)+(2 c d-b e) \left (8 c^2 d^2-8 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}+\frac {\left (d (c d-b e) \left (16 c^2 d^2-16 b c d e-b^2 e^2\right ) \sqrt {x} \sqrt {b+c x}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x} \sqrt {d+e x}} \, dx}{3 b^4 c \sqrt {b x+c x^2}}-\frac {\left (2 (2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) \sqrt {x} \sqrt {b+c x}\right ) \int \frac {\sqrt {d+e x}}{\sqrt {x} \sqrt {b+c x}} \, dx}{3 b^4 c \sqrt {b x+c x^2}}\\ &=-\frac {2 (d+e x)^{5/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (b c d^2 (8 c d-9 b e)+(2 c d-b e) \left (8 c^2 d^2-8 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {\left (2 (2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) \sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {d+e x}\right ) \int \frac {\sqrt {1+\frac {e x}{d}}}{\sqrt {x} \sqrt {1+\frac {c x}{b}}} \, dx}{3 b^4 c \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {\left (d (c d-b e) \left (16 c^2 d^2-16 b c d e-b^2 e^2\right ) \sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {1+\frac {e x}{d}}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {1+\frac {e x}{d}}} \, dx}{3 b^4 c \sqrt {d+e x} \sqrt {b x+c x^2}}\\ &=-\frac {2 (d+e x)^{5/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (b c d^2 (8 c d-9 b e)+(2 c d-b e) \left (8 c^2 d^2-8 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {4 (2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) \sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {d+e x} E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{3 (-b)^{7/2} c^{3/2} \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {2 d (c d-b e) \left (16 c^2 d^2-16 b c d e-b^2 e^2\right ) \sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {1+\frac {e x}{d}} F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{3 (-b)^{7/2} c^{3/2} \sqrt {d+e x} \sqrt {b x+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.14, size = 405, normalized size = 1.06 \[ \frac {2 \left (b (d+e x) \left (-b c d^3 (b+c x)^2+2 c d^2 x (b+c x)^2 (4 c d-5 b e)+b x^2 (c d-b e)^3+2 x^2 (b+c x) (c d-b e)^2 (b e+4 c d)\right )-x \sqrt {\frac {b}{c}} (b+c x) \left (-i b e x^{3/2} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} \left (2 b^3 e^3+3 b^2 c d e^2-13 b c^2 d^2 e+8 c^3 d^3\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+2 i b e x^{3/2} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} \left (b^3 e^3+2 b^2 c d e^2-12 b c^2 d^2 e+8 c^3 d^3\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+2 \sqrt {\frac {b}{c}} (b+c x) (d+e x) \left (b^3 e^3+2 b^2 c d e^2-12 b c^2 d^2 e+8 c^3 d^3\right )\right )\right )}{3 b^5 c (x (b+c x))^{3/2} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(7/2)/(b*x + c*x^2)^(5/2),x]

[Out]

(2*(b*(d + e*x)*(b*(c*d - b*e)^3*x^2 + 2*(c*d - b*e)^2*(4*c*d + b*e)*x^2*(b + c*x) - b*c*d^3*(b + c*x)^2 + 2*c
*d^2*(4*c*d - 5*b*e)*x*(b + c*x)^2) - Sqrt[b/c]*x*(b + c*x)*(2*Sqrt[b/c]*(8*c^3*d^3 - 12*b*c^2*d^2*e + 2*b^2*c
*d*e^2 + b^3*e^3)*(b + c*x)*(d + e*x) + (2*I)*b*e*(8*c^3*d^3 - 12*b*c^2*d^2*e + 2*b^2*c*d*e^2 + b^3*e^3)*Sqrt[
1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] - I*b*e*(8*c^3*d^3
 - 13*b*c^2*d^2*e + 3*b^2*c*d*e^2 + 2*b^3*e^3)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticF[I*ArcSinh
[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])))/(3*b^5*c*(x*(b + c*x))^(3/2)*Sqrt[d + e*x])

________________________________________________________________________________________

fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {e x + d}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, b^{2} c x^{4} + b^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*sqrt(c*x^2 + b*x)*sqrt(e*x + d)/(c^3*x^6 + 3*b*c^2*x^5 + 3*
b^2*c*x^4 + b^3*x^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^(7/2)/(c*x^2 + b*x)^(5/2), x)

________________________________________________________________________________________

maple [B]  time = 0.14, size = 1687, normalized size = 4.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/(c*x^2+b*x)^(5/2),x)

[Out]

2/3*(4*x^4*b^2*c^4*d*e^3+16*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)
/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^2*c^4*d^4-16*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^
(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^2*c^4*d^4+x^3*b^4*c^2*e^4+6*x*b^2*c^4*d^4+16*x^
4*c^6*d^3*e+24*x^2*b*c^5*d^4+2*x^4*b^3*c^3*e^4-33*x^3*b^2*c^4*d^2*e^2-3*x^2*b^3*c^3*d^2*e^2+x^2*b^4*c^2*d*e^3+
2*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e
)^(1/2))*x*b^5*c*d*e^3-28*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b
)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^4*c^2*d^2*e^2+40*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x
)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^3*c^3*d^3*e+((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-
c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x^2*b^4*c^2*d*e^3+15*((c*x
+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2)
)*x^2*b^3*c^3*d^2*e^2-32*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)
^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x^2*b^2*c^4*d^3*e+2*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^
(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x^2*b^4*c^2*d*e^3+((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-
c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^5*c*d*e^3+15*((c*x+b)/
b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*
b^4*c^2*d^2*e^2-28*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b)^(1/2)
,(1/(b*e-c*d)*b*e)^(1/2))*x^2*b^3*c^3*d^2*e^2+40*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/
2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x^2*b^2*c^4*d^3*e-32*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-
c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^3*c^3*d^3*e+16*((c*x+b
)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*
x^2*b*c^5*d^4-31*x^2*b^2*c^4*d^3*e-11*x*b^3*c^3*d^3*e-24*x^4*b*c^5*d^2*e^2+9*x^3*b^3*c^3*d*e^3+16*x^3*c^6*d^4+
2*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e
)^(1/2))*x^2*b^5*c*e^4-16*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b
)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x^2*b*c^5*d^4-b^3*c^3*d^4+2*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-
1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*x*b^6*e^4)/x^2*((c*x+b)*x)^(1/2)/b^4/(c*x+
b)^2/c^3/(e*x+d)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(7/2)/(c*x^2 + b*x)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^{7/2}}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(7/2)/(b*x + c*x^2)^(5/2),x)

[Out]

int((d + e*x)^(7/2)/(b*x + c*x^2)^(5/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(7/2)/(c*x**2+b*x)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________