3.2282 \(\int \frac {\sqrt {d+e x} (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=313 \[ \frac {\sqrt {d+e x} (-2 b e g-c d g+5 c e f)}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {-2 b e g-c d g+5 c e f}{3 c e^2 \sqrt {d+e x} (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {2 \sqrt {d+e x} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(-2 b e g-c d g+5 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{7/2}} \]

[Out]

-(-2*b*e*g-c*d*g+5*c*e*f)*arctanh((d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/(-b*e+2*c*d)^(1/2)/(e*x+d)^(1/2))/e^2
/(-b*e+2*c*d)^(7/2)+2/3*(-b*e*g+c*d*g+c*e*f)*(e*x+d)^(1/2)/c/e^2/(-b*e+2*c*d)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)
^(3/2)+1/3*(2*b*e*g+c*d*g-5*c*e*f)/c/e^2/(-b*e+2*c*d)^2/(e*x+d)^(1/2)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)+(
-2*b*e*g-c*d*g+5*c*e*f)*(e*x+d)^(1/2)/e^2/(-b*e+2*c*d)^3/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {788, 672, 666, 660, 208} \[ \frac {\sqrt {d+e x} (-2 b e g-c d g+5 c e f)}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {-2 b e g-c d g+5 c e f}{3 c e^2 \sqrt {d+e x} (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {2 \sqrt {d+e x} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(-2 b e g-c d g+5 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[d + e*x]*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2),x]

[Out]

(2*(c*e*f + c*d*g - b*e*g)*Sqrt[d + e*x])/(3*c*e^2*(2*c*d - b*e)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))
- (5*c*e*f - c*d*g - 2*b*e*g)/(3*c*e^2*(2*c*d - b*e)^2*Sqrt[d + e*x]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]
) + ((5*c*e*f - c*d*g - 2*b*e*g)*Sqrt[d + e*x])/(e^2*(2*c*d - b*e)^3*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]
) - ((5*c*e*f - c*d*g - 2*b*e*g)*ArcTanh[Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]/(Sqrt[2*c*d - b*e]*Sqrt[d +
e*x])])/(e^2*(2*c*d - b*e)^(7/2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 660

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

Rule 666

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

Rule 672

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

Rule 788

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

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {(5 c e f-c d g-2 b e g) \int \frac {1}{\sqrt {d+e x} \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx}{3 c e (2 c d-b e)}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \int \frac {\sqrt {d+e x}}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx}{2 e (2 c d-b e)^2}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \sqrt {d+e x}}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{2 e (2 c d-b e)^3}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \sqrt {d+e x}}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \operatorname {Subst}\left (\int \frac {1}{-2 c d e^2+b e^3+e^2 x^2} \, dx,x,\frac {\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{\sqrt {d+e x}}\right )}{(2 c d-b e)^3}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \sqrt {d+e x}}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(5 c e f-c d g-2 b e g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {2 c d-b e} \sqrt {d+e x}}\right )}{e^2 (2 c d-b e)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 135, normalized size = 0.43 \[ \frac {(d+e x) (2 b e g+c d g-5 c e f) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {-c d+b e+c e x}{b e-2 c d}\right )-3 (2 c d-b e) (d g-e f)}{3 e^2 \sqrt {d+e x} (b e-2 c d)^2 (b e-c d+c e x) \sqrt {(d+e x) (c (d-e x)-b e)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[d + e*x]*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2),x]

[Out]

(-3*(2*c*d - b*e)*(-(e*f) + d*g) + (-5*c*e*f + c*d*g + 2*b*e*g)*(d + e*x)*Hypergeometric2F1[-3/2, 1, -1/2, (-(
c*d) + b*e + c*e*x)/(-2*c*d + b*e)])/(3*e^2*(-2*c*d + b*e)^2*Sqrt[d + e*x]*(-(c*d) + b*e + c*e*x)*Sqrt[(d + e*
x)*(-(b*e) + c*(d - e*x))])

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fricas [B]  time = 1.09, size = 2106, normalized size = 6.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*((5*c^3*e^5*f - (c^3*d*e^4 + 2*b*c^2*e^5)*g)*x^4 + 2*(5*b*c^2*e^5*f - (b*c^2*d*e^4 + 2*b^2*c*e^5)*g)*x
^3 - (5*(2*c^3*d^2*e^3 - 2*b*c^2*d*e^4 - b^2*c*e^5)*f - (2*c^3*d^3*e^2 + 2*b*c^2*d^2*e^3 - 5*b^2*c*d*e^4 - 2*b
^3*e^5)*g)*x^2 + 5*(c^3*d^4*e - 2*b*c^2*d^3*e^2 + b^2*c*d^2*e^3)*f - (c^3*d^5 - 3*b^2*c*d^3*e^2 + 2*b^3*d^2*e^
3)*g - 2*(5*(b*c^2*d^2*e^3 - b^2*c*d*e^4)*f - (b*c^2*d^3*e^2 + b^2*c*d^2*e^3 - 2*b^3*d*e^4)*g)*x)*sqrt(2*c*d -
b*e)*log(-(c*e^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x + 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)
*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(3
*(5*(2*c^3*d*e^3 - b*c^2*e^4)*f - (2*c^3*d^2*e^2 + 3*b*c^2*d*e^3 - 2*b^2*c*e^4)*g)*x^2 - (26*c^3*d^3*e - 29*b*
c^2*d^2*e^2 + 2*b^2*c*d*e^3 + 3*b^3*e^4)*f - (14*c^3*d^4 - 43*b*c^2*d^3*e + 40*b^2*c*d^2*e^2 - 11*b^3*d*e^3)*g
- 2*(5*(2*c^3*d^2*e^2 - 5*b*c^2*d*e^3 + 2*b^2*c*e^4)*f - (2*c^3*d^3*e - b*c^2*d^2*e^2 - 8*b^2*c*d*e^3 + 4*b^3
*e^4)*g)*x)*sqrt(e*x + d))/(16*c^6*d^8*e^2 - 64*b*c^5*d^7*e^3 + 104*b^2*c^4*d^6*e^4 - 88*b^3*c^3*d^5*e^5 + 41*
b^4*c^2*d^4*e^6 - 10*b^5*c*d^3*e^7 + b^6*d^2*e^8 + (16*c^6*d^4*e^6 - 32*b*c^5*d^3*e^7 + 24*b^2*c^4*d^2*e^8 - 8
*b^3*c^3*d*e^9 + b^4*c^2*e^10)*x^4 + 2*(16*b*c^5*d^4*e^6 - 32*b^2*c^4*d^3*e^7 + 24*b^3*c^3*d^2*e^8 - 8*b^4*c^2
*d*e^9 + b^5*c*e^10)*x^3 - (32*c^6*d^6*e^4 - 96*b*c^5*d^5*e^5 + 96*b^2*c^4*d^4*e^6 - 32*b^3*c^3*d^3*e^7 - 6*b^
4*c^2*d^2*e^8 + 6*b^5*c*d*e^9 - b^6*e^10)*x^2 - 2*(16*b*c^5*d^6*e^4 - 48*b^2*c^4*d^5*e^5 + 56*b^3*c^3*d^4*e^6
- 32*b^4*c^2*d^3*e^7 + 9*b^5*c*d^2*e^8 - b^6*d*e^9)*x), -1/3*(3*((5*c^3*e^5*f - (c^3*d*e^4 + 2*b*c^2*e^5)*g)*x
^4 + 2*(5*b*c^2*e^5*f - (b*c^2*d*e^4 + 2*b^2*c*e^5)*g)*x^3 - (5*(2*c^3*d^2*e^3 - 2*b*c^2*d*e^4 - b^2*c*e^5)*f
- (2*c^3*d^3*e^2 + 2*b*c^2*d^2*e^3 - 5*b^2*c*d*e^4 - 2*b^3*e^5)*g)*x^2 + 5*(c^3*d^4*e - 2*b*c^2*d^3*e^2 + b^2*
c*d^2*e^3)*f - (c^3*d^5 - 3*b^2*c*d^3*e^2 + 2*b^3*d^2*e^3)*g - 2*(5*(b*c^2*d^2*e^3 - b^2*c*d*e^4)*f - (b*c^2*d
^3*e^2 + b^2*c*d^2*e^3 - 2*b^3*d*e^4)*g)*x)*sqrt(-2*c*d + b*e)*arctan(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*
e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(c*e^2*x^2 + b*e^2*x - c*d^2 + b*d*e)) + sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2
- b*d*e)*(3*(5*(2*c^3*d*e^3 - b*c^2*e^4)*f - (2*c^3*d^2*e^2 + 3*b*c^2*d*e^3 - 2*b^2*c*e^4)*g)*x^2 - (26*c^3*d
^3*e - 29*b*c^2*d^2*e^2 + 2*b^2*c*d*e^3 + 3*b^3*e^4)*f - (14*c^3*d^4 - 43*b*c^2*d^3*e + 40*b^2*c*d^2*e^2 - 11*
b^3*d*e^3)*g - 2*(5*(2*c^3*d^2*e^2 - 5*b*c^2*d*e^3 + 2*b^2*c*e^4)*f - (2*c^3*d^3*e - b*c^2*d^2*e^2 - 8*b^2*c*d
*e^3 + 4*b^3*e^4)*g)*x)*sqrt(e*x + d))/(16*c^6*d^8*e^2 - 64*b*c^5*d^7*e^3 + 104*b^2*c^4*d^6*e^4 - 88*b^3*c^3*d
^5*e^5 + 41*b^4*c^2*d^4*e^6 - 10*b^5*c*d^3*e^7 + b^6*d^2*e^8 + (16*c^6*d^4*e^6 - 32*b*c^5*d^3*e^7 + 24*b^2*c^4
*d^2*e^8 - 8*b^3*c^3*d*e^9 + b^4*c^2*e^10)*x^4 + 2*(16*b*c^5*d^4*e^6 - 32*b^2*c^4*d^3*e^7 + 24*b^3*c^3*d^2*e^8
- 8*b^4*c^2*d*e^9 + b^5*c*e^10)*x^3 - (32*c^6*d^6*e^4 - 96*b*c^5*d^5*e^5 + 96*b^2*c^4*d^4*e^6 - 32*b^3*c^3*d^
3*e^7 - 6*b^4*c^2*d^2*e^8 + 6*b^5*c*d*e^9 - b^6*e^10)*x^2 - 2*(16*b*c^5*d^6*e^4 - 48*b^2*c^4*d^5*e^5 + 56*b^3*
c^3*d^4*e^6 - 32*b^4*c^2*d^3*e^7 + 9*b^5*c*d^2*e^8 - b^6*d*e^9)*x)]

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giac [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)^(1/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.08, size = 904, normalized size = 2.89 \[ -\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (6 \sqrt {-c e x -b e +c d}\, b c \,e^{3} g \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+3 \sqrt {-c e x -b e +c d}\, c^{2} d \,e^{2} g \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 \sqrt {-c e x -b e +c d}\, c^{2} e^{3} f \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+6 \sqrt {-c e x -b e +c d}\, b^{2} e^{3} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+3 \sqrt {-c e x -b e +c d}\, b c d \,e^{2} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 \sqrt {-c e x -b e +c d}\, b c \,e^{3} f x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+6 \sqrt {b e -2 c d}\, b c \,e^{3} g \,x^{2}+3 \sqrt {b e -2 c d}\, c^{2} d \,e^{2} g \,x^{2}-15 \sqrt {b e -2 c d}\, c^{2} e^{3} f \,x^{2}+6 \sqrt {-c e x -b e +c d}\, b^{2} d \,e^{2} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+8 \sqrt {b e -2 c d}\, b^{2} e^{3} g x -3 \sqrt {-c e x -b e +c d}\, b c \,d^{2} e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 \sqrt {-c e x -b e +c d}\, b c d \,e^{2} f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-20 \sqrt {b e -2 c d}\, b c \,e^{3} f x -3 \sqrt {-c e x -b e +c d}\, c^{2} d^{3} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+15 \sqrt {-c e x -b e +c d}\, c^{2} d^{2} e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-2 \sqrt {b e -2 c d}\, c^{2} d^{2} e g x +10 \sqrt {b e -2 c d}\, c^{2} d \,e^{2} f x +11 \sqrt {b e -2 c d}\, b^{2} d \,e^{2} g -3 \sqrt {b e -2 c d}\, b^{2} e^{3} f -18 \sqrt {b e -2 c d}\, b c \,d^{2} e g -8 \sqrt {b e -2 c d}\, b c d \,e^{2} f +7 \sqrt {b e -2 c d}\, c^{2} d^{3} g +13 \sqrt {b e -2 c d}\, c^{2} d^{2} e f \right )}{3 \left (e x +d \right )^{\frac {3}{2}} \left (c e x +b e -c d \right )^{2} \left (b e -2 c d \right )^{\frac {7}{2}} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x)

[Out]

-1/3*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*(6*(-c*e*x-b*e+c*d)^(1/2)*b*c*e^3*g*x^2*arctan((-c*e*x-b*e+c*d)^(1
/2)/(b*e-2*c*d)^(1/2))+3*(-c*e*x-b*e+c*d)^(1/2)*c^2*d*e^2*g*x^2*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2
))-15*(-c*e*x-b*e+c*d)^(1/2)*c^2*e^3*f*x^2*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))+6*arctan((-c*e*x-b
*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*x*b^2*e^3*g*(-c*e*x-b*e+c*d)^(1/2)+3*(-c*e*x-b*e+c*d)^(1/2)*b*c*d*e^2*g*x*arc
tan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-15*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*x*b*c*e^3*f*
(-c*e*x-b*e+c*d)^(1/2)+6*(b*e-2*c*d)^(1/2)*b*c*e^3*g*x^2+3*(b*e-2*c*d)^(1/2)*c^2*d*e^2*g*x^2-15*(b*e-2*c*d)^(1
/2)*c^2*e^3*f*x^2+6*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b^2*d*e^2*g*(-c*e*x-b*e+c*d)^(1/2)-3*(-c*
e*x-b*e+c*d)^(1/2)*b*c*d^2*e*g*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-15*arctan((-c*e*x-b*e+c*d)^(1/
2)/(b*e-2*c*d)^(1/2))*b*c*d*e^2*f*(-c*e*x-b*e+c*d)^(1/2)-3*(-c*e*x-b*e+c*d)^(1/2)*c^2*d^3*g*arctan((-c*e*x-b*e
+c*d)^(1/2)/(b*e-2*c*d)^(1/2))+15*(-c*e*x-b*e+c*d)^(1/2)*c^2*d^2*e*f*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)
^(1/2))+8*(b*e-2*c*d)^(1/2)*b^2*e^3*g*x-20*(b*e-2*c*d)^(1/2)*b*c*e^3*f*x-2*(b*e-2*c*d)^(1/2)*c^2*d^2*e*g*x+10*
(b*e-2*c*d)^(1/2)*c^2*d*e^2*f*x+11*(b*e-2*c*d)^(1/2)*b^2*d*e^2*g-3*(b*e-2*c*d)^(1/2)*b^2*e^3*f-18*(b*e-2*c*d)^
(1/2)*b*c*d^2*e*g-8*(b*e-2*c*d)^(1/2)*b*c*d*e^2*f+7*(b*e-2*c*d)^(1/2)*c^2*d^3*g+13*(b*e-2*c*d)^(1/2)*c^2*d^2*e
*f)/(e*x+d)^(3/2)/(c*e*x+b*e-c*d)^2/e^2/(b*e-2*c*d)^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(e*x + d)*(g*x + f)/(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^(1/2))/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2),x)

[Out]

int(((f + g*x)*(d + e*x)^(1/2))/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d + e x} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)*(g*x+f)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(5/2),x)

[Out]

Integral(sqrt(d + e*x)*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(5/2), x)

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