3.13.62 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\) [1262]

Optimal. Leaf size=413 \[ -\frac {2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt {b x+c x^2}}{15 e^4 \sqrt {d+e x}}+\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac {2 \sqrt {-b} \left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {d+e x} E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{15 \sqrt {c} e^5 \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {2 \sqrt {-b} \left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \sqrt {x} \sqrt {1+\frac {c x}{b}} \sqrt {1+\frac {e x}{d}} F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{15 \sqrt {c} e^5 \sqrt {d+e x} \sqrt {b x+c x^2}} \]

[Out]

2/15*(3*B*e*x-5*A*e+8*B*d)*(c*x^2+b*x)^(3/2)/e^2/(e*x+d)^(3/2)-2/15*(40*A*c*e*(-b*e+2*c*d)-B*(3*b^2*e^2-88*b*c
*d*e+128*c^2*d^2))*EllipticE(c^(1/2)*x^(1/2)/(-b)^(1/2),(b*e/c/d)^(1/2))*(-b)^(1/2)*x^(1/2)*(c*x/b+1)^(1/2)*(e
*x+d)^(1/2)/e^5/c^(1/2)/(1+e*x/d)^(1/2)/(c*x^2+b*x)^(1/2)+2/15*(5*A*e*(3*b^2*e^2-16*b*c*d*e+16*c^2*d^2)-B*d*(3
9*b^2*e^2-152*b*c*d*e+128*c^2*d^2))*EllipticF(c^(1/2)*x^(1/2)/(-b)^(1/2),(b*e/c/d)^(1/2))*(-b)^(1/2)*x^(1/2)*(
c*x/b+1)^(1/2)*(1+e*x/d)^(1/2)/e^5/c^(1/2)/(e*x+d)^(1/2)/(c*x^2+b*x)^(1/2)-2/15*(4*B*d*(-9*b*e+16*c*d)-5*A*e*(
-3*b*e+8*c*d)+e*(-10*A*c*e-3*B*b*e+16*B*c*d)*x)*(c*x^2+b*x)^(1/2)/e^4/(e*x+d)^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {826, 857, 729, 113, 111, 118, 117} \begin {gather*} \frac {2 \sqrt {-b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} \left (5 A e \left (3 b^2 e^2-16 b c d e+16 c^2 d^2\right )-B d \left (39 b^2 e^2-152 b c d e+128 c^2 d^2\right )\right ) F\left (\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{15 \sqrt {c} e^5 \sqrt {b x+c x^2} \sqrt {d+e x}}-\frac {2 \sqrt {-b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} \left (40 A c e (2 c d-b e)-B \left (3 b^2 e^2-88 b c d e+128 c^2 d^2\right )\right ) E\left (\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{15 \sqrt {c} e^5 \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}-\frac {2 \sqrt {b x+c x^2} (e x (-10 A c e-3 b B e+16 B c d)-5 A e (8 c d-3 b e)+4 B d (16 c d-9 b e))}{15 e^4 \sqrt {d+e x}}+\frac {2 \left (b x+c x^2\right )^{3/2} (-5 A e+8 B d+3 B e x)}{15 e^2 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(-2*(4*B*d*(16*c*d - 9*b*e) - 5*A*e*(8*c*d - 3*b*e) + e*(16*B*c*d - 3*b*B*e - 10*A*c*e)*x)*Sqrt[b*x + c*x^2])/
(15*e^4*Sqrt[d + e*x]) + (2*(8*B*d - 5*A*e + 3*B*e*x)*(b*x + c*x^2)^(3/2))/(15*e^2*(d + e*x)^(3/2)) - (2*Sqrt[
-b]*(40*A*c*e*(2*c*d - b*e) - B*(128*c^2*d^2 - 88*b*c*d*e + 3*b^2*e^2))*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x
]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*Sqrt[c]*e^5*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x
^2]) + (2*Sqrt[-b]*(5*A*e*(16*c^2*d^2 - 16*b*c*d*e + 3*b^2*e^2) - B*d*(128*c^2*d^2 - 152*b*c*d*e + 39*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)])/(15*
Sqrt[c]*e^5*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

Rule 111

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2*(Sqrt[e]/b)*Rt[-b/
d, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-b/d, 2])], c*(f/(d*e))], 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 113

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 117

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

Rule 118

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 729

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 826

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

Rule 857

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 {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx &=\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac {2 \int \frac {\left (\frac {1}{2} b (8 B d-5 A e)+\frac {1}{2} (16 B c d-3 b B e-10 A c e) x\right ) \sqrt {b x+c x^2}}{(d+e x)^{3/2}} \, dx}{5 e^2}\\ &=-\frac {2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt {b x+c x^2}}{15 e^4 \sqrt {d+e x}}+\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}+\frac {4 \int \frac {\frac {1}{4} b (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e))-\frac {1}{4} \left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) x}{\sqrt {d+e x} \sqrt {b x+c x^2}} \, dx}{15 e^4}\\ &=-\frac {2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt {b x+c x^2}}{15 e^4 \sqrt {d+e x}}+\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac {\left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {b x+c x^2}} \, dx}{15 e^5}+\frac {\left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {b x+c x^2}} \, dx}{15 e^5}\\ &=-\frac {2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt {b x+c x^2}}{15 e^4 \sqrt {d+e x}}+\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac {\left (\left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \sqrt {x} \sqrt {b+c x}\right ) \int \frac {\sqrt {d+e x}}{\sqrt {x} \sqrt {b+c x}} \, dx}{15 e^5 \sqrt {b x+c x^2}}+\frac {\left (\left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \sqrt {x} \sqrt {b+c x}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x} \sqrt {d+e x}} \, dx}{15 e^5 \sqrt {b x+c x^2}}\\ &=-\frac {2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt {b x+c x^2}}{15 e^4 \sqrt {d+e x}}+\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac {\left (\left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\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}{15 e^5 \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {\left (\left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\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}{15 e^5 \sqrt {d+e x} \sqrt {b x+c x^2}}\\ &=-\frac {2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt {b x+c x^2}}{15 e^4 \sqrt {d+e x}}+\frac {2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac {2 \sqrt {-b} \left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\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 )}{15 \sqrt {c} e^5 \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {2 \sqrt {-b} \left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\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 )}{15 \sqrt {c} e^5 \sqrt {d+e x} \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 21.49, size = 436, normalized size = 1.06 \begin {gather*} \frac {2 (x (b+c x))^{3/2} \left (\frac {\left (40 A c e (-2 c d+b e)+B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) (b+c x) (d+e x)}{c}+\frac {e x (b+c x) \left (5 A e \left (-b e (3 d+4 e x)+c \left (8 d^2+10 d e x+e^2 x^2\right )\right )+B \left (b e \left (36 d^2+47 d e x+6 e^2 x^2\right )-c \left (64 d^3+80 d^2 e x+8 d e^2 x^2-3 e^3 x^3\right )\right )\right )}{d+e x}-i \sqrt {\frac {b}{c}} e \left (40 A c e (2 c d-b e)+B \left (-128 c^2 d^2+88 b c d e-3 b^2 e^2\right )\right ) \sqrt {1+\frac {b}{c x}} \sqrt {1+\frac {d}{e x}} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+i \sqrt {\frac {b}{c}} e \left (5 A c e (8 c d-5 b e)+B \left (-64 c^2 d^2+52 b c d e-3 b^2 e^2\right )\right ) \sqrt {1+\frac {b}{c x}} \sqrt {1+\frac {d}{e x}} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )\right )}{15 e^5 x^2 (b+c x)^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(2*(x*(b + c*x))^(3/2)*(((40*A*c*e*(-2*c*d + b*e) + B*(128*c^2*d^2 - 88*b*c*d*e + 3*b^2*e^2))*(b + c*x)*(d + e
*x))/c + (e*x*(b + c*x)*(5*A*e*(-(b*e*(3*d + 4*e*x)) + c*(8*d^2 + 10*d*e*x + e^2*x^2)) + B*(b*e*(36*d^2 + 47*d
*e*x + 6*e^2*x^2) - c*(64*d^3 + 80*d^2*e*x + 8*d*e^2*x^2 - 3*e^3*x^3))))/(d + e*x) - I*Sqrt[b/c]*e*(40*A*c*e*(
2*c*d - b*e) + B*(-128*c^2*d^2 + 88*b*c*d*e - 3*b^2*e^2))*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*Elliptic
E[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] + I*Sqrt[b/c]*e*(5*A*c*e*(8*c*d - 5*b*e) + B*(-64*c^2*d^2 + 52*b*
c*d*e - 3*b^2*e^2))*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)]))/(15*e^5*x^2*(b + c*x)^2*Sqrt[d + e*x])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2366\) vs. \(2(359)=718\).
time = 0.68, size = 2367, normalized size = 5.73

method	result	size
elliptic	\(\frac {\sqrt {x \left (c x +b \right )}\, \sqrt {x \left (e x +d \right ) \left (c x +b \right )}\, \left (\frac {2 d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right ) \sqrt {c e \,x^{3}+b e \,x^{2}+c d \,x^{2}+x b d}}{3 e^{6} \left (x +\frac {d}{e}\right )^{2}}-\frac {2 \left (c e \,x^{2}+b e x \right ) \left (4 A b \,e^{2}-8 A c d e -7 B b d e +11 B c \,d^{2}\right )}{3 e^{5} \sqrt {\left (x +\frac {d}{e}\right ) \left (c e \,x^{2}+b e x \right )}}+\frac {2 B c x \sqrt {c e \,x^{3}+b e \,x^{2}+c d \,x^{2}+x b d}}{5 e^{3}}+\frac {2 \left (\frac {c \left (A c e +2 b B e -2 B c d \right )}{e^{3}}-\frac {2 B c \left (2 b e +2 c d \right )}{5 e^{3}}\right ) \sqrt {c e \,x^{3}+b e \,x^{2}+c d \,x^{2}+x b d}}{3 c e}+\frac {2 \left (\frac {A \,b^{2} e^{3}-4 A b c d \,e^{2}+3 A \,c^{2} d^{2} e -2 B \,b^{2} d \,e^{2}+6 B b c \,d^{2} e -4 B \,c^{2} d^{3}}{e^{5}}+\frac {d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right ) c}{3 e^{5}}-\frac {\left (4 A b \,e^{2}-8 A c d e -7 B b d e +11 B c \,d^{2}\right ) \left (b e -c d \right )}{3 e^{5}}+\frac {b \left (4 A b \,e^{2}-8 A c d e -7 B b d e +11 B c \,d^{2}\right )}{3 e^{4}}-\frac {\left (\frac {c \left (A c e +2 b B e -2 B c d \right )}{e^{3}}-\frac {2 B c \left (2 b e +2 c d \right )}{5 e^{3}}\right ) b d}{3 c e}\right ) b \sqrt {\frac {\left (\frac {b}{c}+x \right ) c}{b}}\, \sqrt {\frac {x +\frac {d}{e}}{-\frac {b}{c}+\frac {d}{e}}}\, \sqrt {-\frac {c x}{b}}\, \EllipticF \left (\sqrt {\frac {\left (\frac {b}{c}+x \right ) c}{b}}, \sqrt {-\frac {b}{c \left (-\frac {b}{c}+\frac {d}{e}\right )}}\right )}{c \sqrt {c e \,x^{3}+b e \,x^{2}+c d \,x^{2}+x b d}}+\frac {2 \left (\frac {2 A b c \,e^{2}-2 A \,c^{2} d e +B \,e^{2} b^{2}-4 e B b c d +3 B \,c^{2} d^{2}}{e^{4}}+\frac {\left (4 A b \,e^{2}-8 A c d e -7 B b d e +11 B c \,d^{2}\right ) c}{3 e^{4}}-\frac {3 B c b d}{5 e^{3}}-\frac {2 \left (\frac {c \left (A c e +2 b B e -2 B c d \right )}{e^{3}}-\frac {2 B c \left (2 b e +2 c d \right )}{5 e^{3}}\right ) \left (b e +c d \right )}{3 c e}\right ) b \sqrt {\frac {\left (\frac {b}{c}+x \right ) c}{b}}\, \sqrt {\frac {x +\frac {d}{e}}{-\frac {b}{c}+\frac {d}{e}}}\, \sqrt {-\frac {c x}{b}}\, \left (\left (-\frac {b}{c}+\frac {d}{e}\right ) \EllipticE \left (\sqrt {\frac {\left (\frac {b}{c}+x \right ) c}{b}}, \sqrt {-\frac {b}{c \left (-\frac {b}{c}+\frac {d}{e}\right )}}\right )-\frac {d \EllipticF \left (\sqrt {\frac {\left (\frac {b}{c}+x \right ) c}{b}}, \sqrt {-\frac {b}{c \left (-\frac {b}{c}+\frac {d}{e}\right )}}\right )}{e}\right )}{c \sqrt {c e \,x^{3}+b e \,x^{2}+c d \,x^{2}+x b d}}\right )}{\sqrt {e x +d}\, x \left (c x +b \right )}\)	\(915\)
default	\(\text {Expression too large to display}\)	\(2367\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*(x*(c*x+b))^(1/2)*(3*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/
b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*e^4*x+128*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*
EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^4+3*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2
)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*d*e^3-128*B*((c*x+b)/b)^(1/2)*(-(e*x+d
)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^4+8*B*c^4*d*e^3
*x^4-50*A*c^4*d*e^3*x^3-6*B*b^2*c^2*e^4*x^3+80*B*c^4*d^2*e^2*x^3-36*B*b^2*c^2*d^2*e^2*x+64*B*b*c^3*d^3*e*x+15*
A*b^2*c^2*d*e^3*x-40*A*b*c^3*d^2*e^2*x-39*B*b*c^3*d*e^3*x^3-35*A*b*c^3*d*e^3*x^2-47*B*b^2*c^2*d*e^3*x^2+44*B*b
*c^3*d^2*e^2*x^2-15*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2
),(b*e/(b*e-c*d))^(1/2))*b^3*c*e^4*x+40*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Ellipt
icE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*e^4*x-15*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(
-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d*e^3+80*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*
c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^2*e^2-80*A*((c*
x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b
*c^3*d^3*e+40*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e
/(b*e-c*d))^(1/2))*b^3*c*d*e^3+64*B*c^4*d^3*e*x^2+20*A*b^2*c^2*e^4*x^2-120*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*
e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^2*e^2+80*A*((c*x+b)/
b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*
d^3*e+39*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e
-c*d))^(1/2))*b^3*c*d^2*e^2-152*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*
x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^3*e-91*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b
)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d^2*e^2+216*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(
b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^3*e+216*B*((c*x+b)
/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c
^2*d^2*e^2*x+80*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b
*e/(b*e-c*d))^(1/2))*b^2*c^2*d*e^3*x-80*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Ellipt
icF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^2*e^2*x-120*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1
/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d*e^3*x+80*A*((c*x+b)/b)^(1/2)*(
-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^2*e^2*x+
39*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))
^(1/2))*b^3*c*d*e^3*x-152*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b
)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^2*e^2*x+128*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)
^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^3*e*x-91*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*
e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d*e^3*x-128*B*((c*x+b)/b
)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d
^3*e*x+15*A*b*c^3*e^4*x^3-3*B*c^4*e^4*x^5-5*A*c^4*e^4*x^4-40*A*c^4*d^2*e^2*x^2-9*B*b*c^3*e^4*x^4)/(c*x+b)/x/(e
*x+d)^(3/2)/c^2/e^5

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/(x*e + d)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.57, size = 843, normalized size = 2.04 \begin {gather*} -\frac {2 \, {\left ({\left (128 \, B c^{3} d^{5} + {\left (3 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2} e^{5} + {\left ({\left (23 \, B b^{2} c + 80 \, A b c^{2}\right )} d x^{2} + 2 \, {\left (3 \, B b^{3} - 5 \, A b^{2} c\right )} d x\right )} e^{4} - {\left (8 \, {\left (19 \, B b c^{2} + 10 \, A c^{3}\right )} d^{2} x^{2} - 2 \, {\left (23 \, B b^{2} c + 80 \, A b c^{2}\right )} d^{2} x - {\left (3 \, B b^{3} - 5 \, A b^{2} c\right )} d^{2}\right )} e^{3} + {\left (128 \, B c^{3} d^{3} x^{2} - 16 \, {\left (19 \, B b c^{2} + 10 \, A c^{3}\right )} d^{3} x + {\left (23 \, B b^{2} c + 80 \, A b c^{2}\right )} d^{3}\right )} e^{2} + 8 \, {\left (32 \, B c^{3} d^{4} x - {\left (19 \, B b c^{2} + 10 \, A c^{3}\right )} d^{4}\right )} e\right )} \sqrt {c} e^{\frac {1}{2}} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (c^{2} d^{2} - b c d e + b^{2} e^{2}\right )} e^{\left (-2\right )}}{3 \, c^{2}}, -\frac {4 \, {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e - 3 \, b^{2} c d e^{2} + 2 \, b^{3} e^{3}\right )} e^{\left (-3\right )}}{27 \, c^{3}}, \frac {{\left (c d + {\left (3 \, c x + b\right )} e\right )} e^{\left (-1\right )}}{3 \, c}\right ) + 3 \, {\left (128 \, B c^{3} d^{4} e + {\left (3 \, B b^{2} c + 40 \, A b c^{2}\right )} x^{2} e^{5} - 2 \, {\left (4 \, {\left (11 \, B b c^{2} + 10 \, A c^{3}\right )} d x^{2} - {\left (3 \, B b^{2} c + 40 \, A b c^{2}\right )} d x\right )} e^{4} + {\left (128 \, B c^{3} d^{2} x^{2} - 16 \, {\left (11 \, B b c^{2} + 10 \, A c^{3}\right )} d^{2} x + {\left (3 \, B b^{2} c + 40 \, A b c^{2}\right )} d^{2}\right )} e^{3} + 8 \, {\left (32 \, B c^{3} d^{3} x - {\left (11 \, B b c^{2} + 10 \, A c^{3}\right )} d^{3}\right )} e^{2}\right )} \sqrt {c} e^{\frac {1}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (c^{2} d^{2} - b c d e + b^{2} e^{2}\right )} e^{\left (-2\right )}}{3 \, c^{2}}, -\frac {4 \, {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e - 3 \, b^{2} c d e^{2} + 2 \, b^{3} e^{3}\right )} e^{\left (-3\right )}}{27 \, c^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (c^{2} d^{2} - b c d e + b^{2} e^{2}\right )} e^{\left (-2\right )}}{3 \, c^{2}}, -\frac {4 \, {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e - 3 \, b^{2} c d e^{2} + 2 \, b^{3} e^{3}\right )} e^{\left (-3\right )}}{27 \, c^{3}}, \frac {{\left (c d + {\left (3 \, c x + b\right )} e\right )} e^{\left (-1\right )}}{3 \, c}\right )\right ) + 3 \, {\left (64 \, B c^{3} d^{3} e^{2} - {\left (3 \, B c^{3} x^{3} - 20 \, A b c^{2} x + {\left (6 \, B b c^{2} + 5 \, A c^{3}\right )} x^{2}\right )} e^{5} + {\left (8 \, B c^{3} d x^{2} + 15 \, A b c^{2} d - {\left (47 \, B b c^{2} + 50 \, A c^{3}\right )} d x\right )} e^{4} + 4 \, {\left (20 \, B c^{3} d^{2} x - {\left (9 \, B b c^{2} + 10 \, A c^{3}\right )} d^{2}\right )} e^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x e + d}\right )}}{45 \, {\left (c^{2} x^{2} e^{8} + 2 \, c^{2} d x e^{7} + c^{2} d^{2} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

-2/45*((128*B*c^3*d^5 + (3*B*b^3 - 5*A*b^2*c)*x^2*e^5 + ((23*B*b^2*c + 80*A*b*c^2)*d*x^2 + 2*(3*B*b^3 - 5*A*b^
2*c)*d*x)*e^4 - (8*(19*B*b*c^2 + 10*A*c^3)*d^2*x^2 - 2*(23*B*b^2*c + 80*A*b*c^2)*d^2*x - (3*B*b^3 - 5*A*b^2*c)
*d^2)*e^3 + (128*B*c^3*d^3*x^2 - 16*(19*B*b*c^2 + 10*A*c^3)*d^3*x + (23*B*b^2*c + 80*A*b*c^2)*d^3)*e^2 + 8*(32
*B*c^3*d^4*x - (19*B*b*c^2 + 10*A*c^3)*d^4)*e)*sqrt(c)*e^(1/2)*weierstrassPInverse(4/3*(c^2*d^2 - b*c*d*e + b^
2*e^2)*e^(-2)/c^2, -4/27*(2*c^3*d^3 - 3*b*c^2*d^2*e - 3*b^2*c*d*e^2 + 2*b^3*e^3)*e^(-3)/c^3, 1/3*(c*d + (3*c*x
+ b)*e)*e^(-1)/c) + 3*(128*B*c^3*d^4*e + (3*B*b^2*c + 40*A*b*c^2)*x^2*e^5 - 2*(4*(11*B*b*c^2 + 10*A*c^3)*d*x^
2 - (3*B*b^2*c + 40*A*b*c^2)*d*x)*e^4 + (128*B*c^3*d^2*x^2 - 16*(11*B*b*c^2 + 10*A*c^3)*d^2*x + (3*B*b^2*c + 4
0*A*b*c^2)*d^2)*e^3 + 8*(32*B*c^3*d^3*x - (11*B*b*c^2 + 10*A*c^3)*d^3)*e^2)*sqrt(c)*e^(1/2)*weierstrassZeta(4/
3*(c^2*d^2 - b*c*d*e + b^2*e^2)*e^(-2)/c^2, -4/27*(2*c^3*d^3 - 3*b*c^2*d^2*e - 3*b^2*c*d*e^2 + 2*b^3*e^3)*e^(-
3)/c^3, weierstrassPInverse(4/3*(c^2*d^2 - b*c*d*e + b^2*e^2)*e^(-2)/c^2, -4/27*(2*c^3*d^3 - 3*b*c^2*d^2*e - 3
*b^2*c*d*e^2 + 2*b^3*e^3)*e^(-3)/c^3, 1/3*(c*d + (3*c*x + b)*e)*e^(-1)/c)) + 3*(64*B*c^3*d^3*e^2 - (3*B*c^3*x^
3 - 20*A*b*c^2*x + (6*B*b*c^2 + 5*A*c^3)*x^2)*e^5 + (8*B*c^3*d*x^2 + 15*A*b*c^2*d - (47*B*b*c^2 + 50*A*c^3)*d*
x)*e^4 + 4*(20*B*c^3*d^2*x - (9*B*b*c^2 + 10*A*c^3)*d^2)*e^3)*sqrt(c*x^2 + b*x)*sqrt(x*e + d))/(c^2*x^2*e^8 +
2*c^2*d*x*e^7 + c^2*d^2*e^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/(d + e*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/(x*e + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x)

[Out]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/(d + e*x)^(5/2), x)

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