3.21.51 \(\int \frac {(d+e x)^{3/2} (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=221 \[ \frac {2 (d+e x)^{3/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} (e f-d g)}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{5/2}} \]

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Rubi [A]  time = 0.30, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {788, 666, 660, 208} \begin {gather*} \frac {2 (d+e x)^{3/2} (-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 \sqrt {d+e x} (e f-d g)}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 (e f-d g) \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)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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 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 {(d+e x)^{3/2} (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) (d+e x)^{3/2}}{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 {(e f-d 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}{e (2 c d-b e)}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^{3/2}}{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 (e f-d g) \sqrt {d+e x}}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(e f-d 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}{e (2 c d-b e)^2}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^{3/2}}{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 (e f-d g) \sqrt {d+e x}}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(2 (e f-d 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)^2}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^{3/2}}{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 (e f-d g) \sqrt {d+e x}}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 (e f-d 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)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 146, normalized size = 0.66 \begin {gather*} \frac {2 \sqrt {d+e x} \left (3 c (e f-d g) (b e-c d+c e x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {-c d+b e+c e x}{b e-2 c d}\right )-(2 c d-b e) (-b e g+c d g+c e f)\right )}{3 c e^2 (b e-2 c d)^2 (b e-c d+c e x) \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(-((2*c*d - b*e)*(c*e*f + c*d*g - b*e*g)) + 3*c*(e*f - d*g)*(-(c*d) + b*e + c*e*x)*Hypergeome
tric2F1[-1/2, 1, 1/2, (-(c*d) + b*e + c*e*x)/(-2*c*d + b*e)]))/(3*c*e^2*(-2*c*d + b*e)^2*(-(c*d) + b*e + c*e*x
)*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

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IntegrateAlgebraic [A]  time = 8.77, size = 252, normalized size = 1.14 \begin {gather*} \frac {2 \left (b^2 e^2 g (d+e x)^{3/2}-4 b c e^2 f (d+e x)^{3/2}-4 c^2 d^2 g (d+e x)^{3/2}-3 c^2 e f (d+e x)^{5/2}+8 c^2 d e f (d+e x)^{3/2}+3 c^2 d g (d+e x)^{5/2}\right )}{3 c e^2 (2 c d-b e)^2 \left (-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)\right )^{3/2}}-\frac {2 (d g-e f) \tan ^{-1}\left (\frac {\sqrt {b e-2 c d} \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2}}{\sqrt {d+e x} (b e+c (d+e x)-2 c d)}\right )}{e^2 (b e-2 c d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(8*c^2*d*e*f*(d + e*x)^(3/2) - 4*b*c*e^2*f*(d + e*x)^(3/2) - 4*c^2*d^2*g*(d + e*x)^(3/2) + b^2*e^2*g*(d + e
*x)^(3/2) - 3*c^2*e*f*(d + e*x)^(5/2) + 3*c^2*d*g*(d + e*x)^(5/2)))/(3*c*e^2*(2*c*d - b*e)^2*(2*c*d*(d + e*x)
- b*e*(d + e*x) - c*(d + e*x)^2)^(3/2)) - (2*(-(e*f) + d*g)*ArcTan[(Sqrt[-2*c*d + b*e]*Sqrt[(2*c*d - b*e)*(d +
e*x) - c*(d + e*x)^2])/(Sqrt[d + e*x]*(-2*c*d + b*e + c*(d + e*x)))])/(e^2*(-2*c*d + b*e)^(5/2))

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fricas [B]  time = 0.44, size = 1437, normalized size = 6.50

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*((c^3*e^4*f - c^3*d*e^3*g)*x^3 - ((c^3*d*e^3 - 2*b*c^2*e^4)*f - (c^3*d^2*e^2 - 2*b*c^2*d*e^3)*g)*x^2
+ (c^3*d^3*e - 2*b*c^2*d^2*e^2 + b^2*c*d*e^3)*f - (c^3*d^4 - 2*b*c^2*d^3*e + b^2*c*d^2*e^2)*g - ((c^3*d^2*e^2
- b^2*c*e^4)*f - (c^3*d^3*e - b^2*c*d*e^3)*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)*((10*c^3*d^2*e - 13*b*c^2*d*e^2 + 4*b^2*c*e^3)*f -
(2*c^3*d^3 - b*c^2*d^2*e - 2*b^2*c*d*e^2 + b^3*e^3)*g - 3*((2*c^3*d*e^2 - b*c^2*e^3)*f - (2*c^3*d^2*e - b*c^2*
d*e^2)*g)*x)*sqrt(e*x + d))/(8*c^6*d^6*e^2 - 28*b*c^5*d^5*e^3 + 38*b^2*c^4*d^4*e^4 - 25*b^3*c^3*d^3*e^5 + 8*b^
4*c^2*d^2*e^6 - b^5*c*d*e^7 + (8*c^6*d^3*e^5 - 12*b*c^5*d^2*e^6 + 6*b^2*c^4*d*e^7 - b^3*c^3*e^8)*x^3 - (8*c^6*
d^4*e^4 - 28*b*c^5*d^3*e^5 + 30*b^2*c^4*d^2*e^6 - 13*b^3*c^3*d*e^7 + 2*b^4*c^2*e^8)*x^2 - (8*c^6*d^5*e^3 - 12*
b*c^5*d^4*e^4 - 2*b^2*c^4*d^3*e^5 + 11*b^3*c^3*d^2*e^6 - 6*b^4*c^2*d*e^7 + b^5*c*e^8)*x), -2/3*(3*((c^3*e^4*f
- c^3*d*e^3*g)*x^3 - ((c^3*d*e^3 - 2*b*c^2*e^4)*f - (c^3*d^2*e^2 - 2*b*c^2*d*e^3)*g)*x^2 + (c^3*d^3*e - 2*b*c^
2*d^2*e^2 + b^2*c*d*e^3)*f - (c^3*d^4 - 2*b*c^2*d^3*e + b^2*c*d^2*e^2)*g - ((c^3*d^2*e^2 - b^2*c*e^4)*f - (c^3
*d^3*e - b^2*c*d*e^3)*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)*((10*c
^3*d^2*e - 13*b*c^2*d*e^2 + 4*b^2*c*e^3)*f - (2*c^3*d^3 - b*c^2*d^2*e - 2*b^2*c*d*e^2 + b^3*e^3)*g - 3*((2*c^3
*d*e^2 - b*c^2*e^3)*f - (2*c^3*d^2*e - b*c^2*d*e^2)*g)*x)*sqrt(e*x + d))/(8*c^6*d^6*e^2 - 28*b*c^5*d^5*e^3 + 3
8*b^2*c^4*d^4*e^4 - 25*b^3*c^3*d^3*e^5 + 8*b^4*c^2*d^2*e^6 - b^5*c*d*e^7 + (8*c^6*d^3*e^5 - 12*b*c^5*d^2*e^6 +
6*b^2*c^4*d*e^7 - b^3*c^3*e^8)*x^3 - (8*c^6*d^4*e^4 - 28*b*c^5*d^3*e^5 + 30*b^2*c^4*d^2*e^6 - 13*b^3*c^3*d*e^
7 + 2*b^4*c^2*e^8)*x^2 - (8*c^6*d^5*e^3 - 12*b*c^5*d^4*e^4 - 2*b^2*c^4*d^3*e^5 + 11*b^3*c^3*d^2*e^6 - 6*b^4*c^
2*d*e^7 + b^5*c*e^8)*x)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(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 = 0.07, size = 485, normalized size = 2.19 \begin {gather*} \frac {2 \left (3 \sqrt {-c e x -b e +c d}\, c^{2} d e 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}\, c^{2} e^{2} f 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 g \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 \,e^{2} f \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^{2} g \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 f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+3 \sqrt {b e -2 c d}\, c^{2} d e g x -3 \sqrt {b e -2 c d}\, c^{2} e^{2} f x +\sqrt {b e -2 c d}\, b^{2} e^{2} g -4 \sqrt {b e -2 c d}\, b c \,e^{2} f -\sqrt {b e -2 c d}\, c^{2} d^{2} g +5 \sqrt {b e -2 c d}\, c^{2} d e f \right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}{3 \left (b e -2 c d \right )^{\frac {5}{2}} \left (c e x +b e -c d \right )^{2} \sqrt {e x +d}\, c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x)

[Out]

2/3*(3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*x*c^2*d*e*g*(-c*e*x-b*e+c*d)^(1/2)-3*arctan((-c*e*x-b*
e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*x*c^2*e^2*f*(-c*e*x-b*e+c*d)^(1/2)+3*(-c*e*x-b*e+c*d)^(1/2)*arctan((-c*e*x-b*e
+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d*e*g-3*(-c*e*x-b*e+c*d)^(1/2)*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1
/2))*b*c*e^2*f-3*(-c*e*x-b*e+c*d)^(1/2)*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d^2*g+3*(-c*e*x-b
*e+c*d)^(1/2)*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d*e*f+3*(b*e-2*c*d)^(1/2)*x*c^2*d*e*g-3*(b*
e-2*c*d)^(1/2)*x*c^2*e^2*f+(b*e-2*c*d)^(1/2)*b^2*e^2*g-4*(b*e-2*c*d)^(1/2)*b*c*e^2*f-(b*e-2*c*d)^(1/2)*c^2*d^2
*g+5*(b*e-2*c*d)^(1/2)*c^2*d*e*f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(b*e-2*c*d)^(5/2)/e^2/c/(c*e*x+b*e-c*
d)^2/(e*x+d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(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((e*x + d)^(3/2)*(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 \begin {gather*} \int \frac {\left (f+g\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^(3/2))/(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)^(3/2))/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(g*x+f)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(5/2),x)

[Out]

Timed out

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