3.1255 \(\int (A+B x) \sqrt {d+e x} \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=433 \[ \frac {2 \sqrt {b x+c x^2} \sqrt {d+e x} \left (3 c e x (7 A c e-4 b B e+B c d)+7 A c e (b e+c d)-B \left (4 b^2 e^2-2 b c d e+4 c^2 d^2\right )\right )}{105 c^2 e^2}+\frac {2 \sqrt {-b} d \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (c d-b e) \left (7 A c e (2 c d-b e)-B \left (-4 b^2 e^2-b c d e+8 c^2 d^2\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{105 c^{5/2} e^3 \sqrt {b x+c x^2} \sqrt {d+e x}}+\frac {2 \sqrt {-b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} \left (\left (-2 b^2 e^2-3 b c d e+8 c^2 d^2\right ) (7 A c e-4 b B e+B c d)+5 c d e (3 b B-7 A c) (2 c d-b e)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{105 c^{5/2} e^3 \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}+\frac {2 B \left (b x+c x^2\right )^{3/2} \sqrt {d+e x}}{7 c} \]

[Out]

2/7*B*(c*x^2+b*x)^(3/2)*(e*x+d)^(1/2)/c+2/105*(5*c*(-7*A*c+3*B*b)*d*e*(-b*e+2*c*d)+(7*A*c*e-4*B*b*e+B*c*d)*(-2
*b^2*e^2-3*b*c*d*e+8*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)/c^(5/2)/e^3/(1+e*x/d)^(1/2)/(c*x^2+b*x)^(1/2)+2/105*d*(-b*e+c*d)*(7*A*c*e*(-b*e+2*c*d)
-B*(-4*b^2*e^2-b*c*d*e+8*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)/c^(5/2)/e^3/(e*x+d)^(1/2)/(c*x^2+b*x)^(1/2)+2/105*(7*A*c*e*(b*e+c*d)-B*(4*b^2*e^
2-2*b*c*d*e+4*c^2*d^2)+3*c*e*(7*A*c*e-4*B*b*e+B*c*d)*x)*(e*x+d)^(1/2)*(c*x^2+b*x)^(1/2)/c^2/e^2

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Rubi [A]  time = 0.60, antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {832, 814, 843, 715, 112, 110, 117, 116} \[ \frac {2 \sqrt {b x+c x^2} \sqrt {d+e x} \left (3 c e x (7 A c e-4 b B e+B c d)+7 A c e (b e+c d)-B \left (4 b^2 e^2-2 b c d e+4 c^2 d^2\right )\right )}{105 c^2 e^2}+\frac {2 \sqrt {-b} d \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (c d-b e) \left (7 A c e (2 c d-b e)-B \left (-4 b^2 e^2-b c d e+8 c^2 d^2\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{105 c^{5/2} e^3 \sqrt {b x+c x^2} \sqrt {d+e x}}+\frac {2 \sqrt {-b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} \left (\left (-2 b^2 e^2-3 b c d e+8 c^2 d^2\right ) (7 A c e-4 b B e+B c d)+5 c d e (3 b B-7 A c) (2 c d-b e)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{105 c^{5/2} e^3 \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}+\frac {2 B \left (b x+c x^2\right )^{3/2} \sqrt {d+e x}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(7*A*c*e*(c*d + b*e) - B*(4*c^2*d^2 - 2*b*c*d*e + 4*b^2*e^2) + 3*c*e*(B*c*d - 4*b*B*e + 7*A*c
*e)*x)*Sqrt[b*x + c*x^2])/(105*c^2*e^2) + (2*B*Sqrt[d + e*x]*(b*x + c*x^2)^(3/2))/(7*c) + (2*Sqrt[-b]*(5*c*(3*
b*B - 7*A*c)*d*e*(2*c*d - b*e) + (B*c*d - 4*b*B*e + 7*A*c*e)*(8*c^2*d^2 - 3*b*c*d*e - 2*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)])/(105*c^(5/2)*e^3*Sqrt[
1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (2*Sqrt[-b]*d*(c*d - b*e)*(7*A*c*e*(2*c*d - b*e) - B*(8*c^2*d^2 - b*c*d*e -
4*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)])/(105*c^(5/2)*e^3*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 814

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

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 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 (A+B x) \sqrt {d+e x} \sqrt {b x+c x^2} \, dx &=\frac {2 B \sqrt {d+e x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {2 \int \frac {\left (-\frac {1}{2} (3 b B-7 A c) d+\frac {1}{2} (B c d-4 b B e+7 A c e) x\right ) \sqrt {b x+c x^2}}{\sqrt {d+e x}} \, dx}{7 c}\\ &=\frac {2 \sqrt {d+e x} \left (7 A c e (c d+b e)-B \left (4 c^2 d^2-2 b c d e+4 b^2 e^2\right )+3 c e (B c d-4 b B e+7 A c e) x\right ) \sqrt {b x+c x^2}}{105 c^2 e^2}+\frac {2 B \sqrt {d+e x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 \int \frac {\frac {1}{4} b d \left (7 A c e (c d+b e)-B \left (4 c^2 d^2-2 b c d e+4 b^2 e^2\right )\right )-\frac {1}{4} \left (5 c (3 b B-7 A c) d e (2 c d-b e)+2 (B c d-4 b B e+7 A c e) \left (4 c^2 d^2-\frac {3}{2} b c d e-b^2 e^2\right )\right ) x}{\sqrt {d+e x} \sqrt {b x+c x^2}} \, dx}{105 c^2 e^2}\\ &=\frac {2 \sqrt {d+e x} \left (7 A c e (c d+b e)-B \left (4 c^2 d^2-2 b c d e+4 b^2 e^2\right )+3 c e (B c d-4 b B e+7 A c e) x\right ) \sqrt {b x+c x^2}}{105 c^2 e^2}+\frac {2 B \sqrt {d+e x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {\left (d (c d-b e) \left (7 A c e (2 c d-b e)-B \left (8 c^2 d^2-b c d e-4 b^2 e^2\right )\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {b x+c x^2}} \, dx}{105 c^2 e^3}+\frac {\left (5 c (3 b B-7 A c) d e (2 c d-b e)+(B c d-4 b B e+7 A c e) \left (8 c^2 d^2-3 b c d e-2 b^2 e^2\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {b x+c x^2}} \, dx}{105 c^2 e^3}\\ &=\frac {2 \sqrt {d+e x} \left (7 A c e (c d+b e)-B \left (4 c^2 d^2-2 b c d e+4 b^2 e^2\right )+3 c e (B c d-4 b B e+7 A c e) x\right ) \sqrt {b x+c x^2}}{105 c^2 e^2}+\frac {2 B \sqrt {d+e x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {\left (d (c d-b e) \left (7 A c e (2 c d-b e)-B \left (8 c^2 d^2-b c d e-4 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}{105 c^2 e^3 \sqrt {b x+c x^2}}+\frac {\left (\left (5 c (3 b B-7 A c) d e (2 c d-b e)+(B c d-4 b B e+7 A c e) \left (8 c^2 d^2-3 b c d e-2 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}{105 c^2 e^3 \sqrt {b x+c x^2}}\\ &=\frac {2 \sqrt {d+e x} \left (7 A c e (c d+b e)-B \left (4 c^2 d^2-2 b c d e+4 b^2 e^2\right )+3 c e (B c d-4 b B e+7 A c e) x\right ) \sqrt {b x+c x^2}}{105 c^2 e^2}+\frac {2 B \sqrt {d+e x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {\left (\left (5 c (3 b B-7 A c) d e (2 c d-b e)+(B c d-4 b B e+7 A c e) \left (8 c^2 d^2-3 b c d e-2 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}{105 c^2 e^3 \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {\left (d (c d-b e) \left (7 A c e (2 c d-b e)-B \left (8 c^2 d^2-b c d e-4 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}{105 c^2 e^3 \sqrt {d+e x} \sqrt {b x+c x^2}}\\ &=\frac {2 \sqrt {d+e x} \left (7 A c e (c d+b e)-B \left (4 c^2 d^2-2 b c d e+4 b^2 e^2\right )+3 c e (B c d-4 b B e+7 A c e) x\right ) \sqrt {b x+c x^2}}{105 c^2 e^2}+\frac {2 B \sqrt {d+e x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {2 \sqrt {-b} \left (5 c (3 b B-7 A c) d e (2 c d-b e)+(B c d-4 b B e+7 A c e) \left (8 c^2 d^2-3 b c d e-2 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 )}{105 c^{5/2} e^3 \sqrt {1+\frac {e x}{d}} \sqrt {b x+c x^2}}+\frac {2 \sqrt {-b} d (c d-b e) \left (7 A c e (2 c d-b e)-B \left (8 c^2 d^2-b c d e-4 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 )}{105 c^{5/2} e^3 \sqrt {d+e x} \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 3.37, size = 461, normalized size = 1.06 \[ -\frac {2 \left (b e x (b+c x) (d+e x) \left (B \left (4 b^2 e^2-b c e (2 d+3 e x)+c^2 \left (4 d^2-3 d e x-15 e^2 x^2\right )\right )-7 A c e (b e+c (d+3 e x))\right )+\sqrt {\frac {b}{c}} \left (-i b e x^{3/2} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} (c d-b e) \left (7 A c e (c d-2 b e)-B \left (-8 b^2 e^2+b c d e+4 c^2 d^2\right )\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+i b e x^{3/2} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} \left (14 A c e \left (b^2 e^2-b c d e+c^2 d^2\right )+B \left (-8 b^3 e^3+5 b^2 c d e^2+5 b c^2 d^2 e-8 c^3 d^3\right )\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+\sqrt {\frac {b}{c}} (b+c x) (d+e x) \left (14 A c e \left (b^2 e^2-b c d e+c^2 d^2\right )+B \left (-8 b^3 e^3+5 b^2 c d e^2+5 b c^2 d^2 e-8 c^3 d^3\right )\right )\right )\right )}{105 b c^2 e^3 \sqrt {x (b+c x)} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*e*x*(b + c*x)*(d + e*x)*(-7*A*c*e*(b*e + c*(d + 3*e*x)) + B*(4*b^2*e^2 - b*c*e*(2*d + 3*e*x) + c^2*(4*d
^2 - 3*d*e*x - 15*e^2*x^2))) + Sqrt[b/c]*(Sqrt[b/c]*(14*A*c*e*(c^2*d^2 - b*c*d*e + b^2*e^2) + B*(-8*c^3*d^3 +
5*b*c^2*d^2*e + 5*b^2*c*d*e^2 - 8*b^3*e^3))*(b + c*x)*(d + e*x) + I*b*e*(14*A*c*e*(c^2*d^2 - b*c*d*e + b^2*e^2
) + B*(-8*c^3*d^3 + 5*b*c^2*d^2*e + 5*b^2*c*d*e^2 - 8*b^3*e^3))*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*El
lipticE[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] - I*b*e*(c*d - b*e)*(7*A*c*e*(c*d - 2*b*e) - B*(4*c^2*d^2 +
b*c*d*e - 8*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)])))/(105*b*c^2*e^3*Sqrt[x*(b + c*x)]*Sqrt[d + e*x])

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{2} + b x} {\left (B x + A\right )} \sqrt {e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x)*(B*x + A)*sqrt(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + b*x)*(B*x + A)*sqrt(e*x + d), x)

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maple [B]  time = 0.14, size = 1526, normalized size = 3.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(1/2)*(c*x^2+b*x)^(1/2),x)

[Out]

2/105*(e*x+d)^(1/2)*((c*x+b)*x)^(1/2)*(14*A*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(
1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*b^4*c*e^4-4*B*x*b*c^4*d^3*e+7*A*x^2*c^5*d^2*e^2-4*B*x^2*b^3
*c^2*e^4-4*B*x^2*c^5*d^3*e+18*B*x^4*b*c^4*e^4+18*B*x^4*c^5*d*e^3+28*A*x^3*b*c^4*e^4+28*A*x^3*c^5*d*e^3-B*x^3*b
^2*c^3*e^4-B*x^3*c^5*d^2*e^2+7*A*x^2*b^2*c^3*e^4-13*B*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c
*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*b^2*c^3*d^3*e-4*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(
b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b^4*c*d*e^3+3*B*((c*x+b)
/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b
^3*c^2*d^2*e^2+8*B*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d
))^(1/2)*(-1/b*c*x)^(1/2)*b*c^4*d^4-8*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*Ellipt
icF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b*c^4*d^4+9*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-
1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b^2*c^3*d^3*e-28*A*EllipticE(((c*x+b)/b)^(
1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*b^3*c^2*d*e^3+28
*A*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c
*x)^(1/2)*b^2*c^3*d^2*e^2-14*A*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d
)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*b*c^4*d^3*e+7*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x
)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b^3*c^2*d*e^3-21*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/
(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b^2*c^3*d^2*e^2+14*A*((
c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1
/2))*b*c^4*d^3*e+13*B*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-
c*d))^(1/2)*(-1/b*c*x)^(1/2)*b^4*c*d*e^3-4*B*x*b^3*c^2*d*e^3+2*B*x*b^2*c^3*d^2*e^2+7*A*x*b^2*c^3*d*e^3+7*A*x*b
*c^4*d^2*e^2+B*x^2*b^2*c^3*d*e^3+B*x^2*b*c^4*d^2*e^2+35*A*x^2*b*c^4*d*e^3+23*B*x^3*b*c^4*d*e^3+15*B*x^5*c^5*e^
4+21*A*x^4*c^5*e^4-8*B*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e
-c*d))^(1/2)*(-1/b*c*x)^(1/2)*b^5*e^4)/x/(c*e*x^2+b*e*x+c*d*x+b*d)/c^4/e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)*(B*x + A)*sqrt(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)*(A + B*x)*(d + e*x)^(1/2),x)

[Out]

int((b*x + c*x^2)^(1/2)*(A + B*x)*(d + e*x)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x \left (b + c x\right )} \left (A + B x\right ) \sqrt {d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(1/2)*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)*sqrt(d + e*x), x)

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