3.1617 \(\int \frac{b+2 c x}{(d+e x)^{5/2} \left (a+b x+c x^2\right )} \, dx\)

Optimal. Leaf size=518 \[ \frac{2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{2} \sqrt{c} \left (2 c^2 d \left (d \sqrt{b^2-4 a c}+4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}+b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (-2 c^2 d \left (d \sqrt{b^2-4 a c}-4 a e\right )-2 c e \left (-b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (b-\sqrt{b^2-4 a c}\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{2 (2 c d-b e)}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )} \]

[Out]

(2*(2*c*d - b*e))/(3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (2*(2*c^2*d^2 +
b^2*e^2 - 2*c*e*(b*d + a*e)))/((c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x]) - (Sqrt[
2]*Sqrt[c]*(b^2*(b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c^2*d*(Sqrt[b^2 - 4*a*c]*d + 4*a
*e) - 2*c*e*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*A
rcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])
/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e
^2)^2) + (Sqrt[2]*Sqrt[c]*(b^2*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c^2*d*(Sqrt[b^2 -
4*a*c]*d - 4*a*e) - 2*c*e*(b^2*d - b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e - a*Sqrt[b^2
- 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2
- 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^
2 - b*d*e + a*e^2)^2)

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Rubi [A]  time = 4.42499, antiderivative size = 518, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143 \[ \frac{2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{2} \sqrt{c} \left (2 c^2 d \left (d \sqrt{b^2-4 a c}+4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}+b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (-2 c^2 d \left (d \sqrt{b^2-4 a c}-4 a e\right )-2 c e \left (-b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (b-\sqrt{b^2-4 a c}\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{2 (2 c d-b e)}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

(2*(2*c*d - b*e))/(3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (2*(2*c^2*d^2 +
b^2*e^2 - 2*c*e*(b*d + a*e)))/((c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x]) - (Sqrt[
2]*Sqrt[c]*(b^2*(b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c^2*d*(Sqrt[b^2 - 4*a*c]*d + 4*a
*e) - 2*c*e*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*A
rcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])
/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e
^2)^2) + (Sqrt[2]*Sqrt[c]*(b^2*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c^2*d*(Sqrt[b^2 -
4*a*c]*d - 4*a*e) - 2*c*e*(b^2*d - b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e - a*Sqrt[b^2
- 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2
- 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^
2 - b*d*e + a*e^2)^2)

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((2*c*x+b)/(e*x+d)**(5/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

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Mathematica [A]  time = 2.2206, size = 480, normalized size = 0.93 \[ \frac{\frac{6 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\sqrt{d+e x}}-\frac{3 \sqrt{2} \sqrt{c} \left (2 c^2 d \left (d \sqrt{b^2-4 a c}+4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}+b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{b^2-4 a c} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{3 \sqrt{2} \sqrt{c} \left (2 c^2 d \left (d \sqrt{b^2-4 a c}-4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}-2 a b e+b^2 (-d)\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}-b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{2 (2 c d-b e) \left (e (a e-b d)+c d^2\right )}{(d+e x)^{3/2}}}{3 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

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

[Out]

((2*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e)))/(d + e*x)^(3/2) + (6*(2*c^2*d^2 +
b^2*e^2 - 2*c*e*(b*d + a*e)))/Sqrt[d + e*x] - (3*Sqrt[2]*Sqrt[c]*(b^2*(b + Sqrt[
b^2 - 4*a*c])*e^2 + 2*c^2*d*(Sqrt[b^2 - 4*a*c]*d + 4*a*e) - 2*c*e*(b^2*d + b*Sqr
t[b^2 - 4*a*c]*d + 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sq
rt[d + e*x])/Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2
*c*d + (-b + Sqrt[b^2 - 4*a*c])*e]) - (3*Sqrt[2]*Sqrt[c]*(b^2*(-b + Sqrt[b^2 - 4
*a*c])*e^2 + 2*c^2*d*(Sqrt[b^2 - 4*a*c]*d - 4*a*e) - 2*c*e*(-(b^2*d) + b*Sqrt[b^
2 - 4*a*c]*d - 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d
+ e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d
- (b + Sqrt[b^2 - 4*a*c])*e]))/(3*(c*d^2 + e*(-(b*d) + a*e))^2)

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Maple [B]  time = 0.056, size = 1962, normalized size = 3.8 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((2*c*x+b)/(e*x+d)^(5/2)/(c*x^2+b*x+a),x)

[Out]

-2/3/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)*b*e+4/3/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)
*c*d-4/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^(1/2)*a*c*e^2+2/(a*e^2-b*d*e+c*d^2)^2/(e*x+
d)^(1/2)*b^2*e^2-4/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^(1/2)*b*c*d*e+4/(a*e^2-b*d*e+c*
d^2)^2/(e*x+d)^(1/2)*c^2*d^2+4/(a*e^2-b*d*e+c*d^2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2
)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2
)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*b*e^3-8/(a*e^2-b*d*
e+c*d^2)^2*c^3/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^
(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^
(1/2))*c)^(1/2))*a*e^2*d-1/(a*e^2-b*d*e+c*d^2)^2*c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1
/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)*2^(1
/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^3*e^3+2/(a*e^2-b*d*e+c*d^
2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))
*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))
*c)^(1/2))*b^2*d*e^2+2/(a*e^2-b*d*e+c*d^2)^2*c^2*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a
*c-b^2))^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a
*c-b^2))^(1/2))*c)^(1/2))*a*e^2-1/(a*e^2-b*d*e+c*d^2)^2*c*2^(1/2)/((-b*e+2*c*d+(
-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)*2^(1/2)/((-b*e+2*c*d+(
-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*e^2+2/(a*e^2-b*d*e+c*d^2)^2*c^2*2^(1/2)/(
(-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)*2^(1/2)/(
(-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d*e-2/(a*e^2-b*d*e+c*d^2)^2*c^
3*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2
)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^2+4/(a*e^2-b*d*e+c*
d^2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2)
)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*
c)^(1/2))*a*b*e^3-8/(a*e^2-b*d*e+c*d^2)^2*c^3/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/(
(b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(c*(e*x+d)^(1/2)*2^(1/2)/((b
*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*e^2*d-1/(a*e^2-b*d*e+c*d^2)^2*c/(
-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*a
rctan(c*(e*x+d)^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^
3*e^3+2/(a*e^2-b*d*e+c*d^2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(
-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(c*(e*x+d)^(1/2)*2^(1/2)/((b*e-2*c*d+(-e
^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d*e^2-2/(a*e^2-b*d*e+c*d^2)^2*c^2*2^(1/2)/(
(b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(c*(e*x+d)^(1/2)*2^(1/2)/((b
*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*e^2+1/(a*e^2-b*d*e+c*d^2)^2*c*2^(
1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(c*(e*x+d)^(1/2)*2^(1/
2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*e^2-2/(a*e^2-b*d*e+c*d^2)
^2*c^2*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(c*(e*x+d)^(
1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d*e+2/(a*e^2-b*d*
e+c*d^2)^2*c^3*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(c*(
e*x+d)^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x + b)/((c*x^2 + b*x + a)*(e*x + d)^(5/2)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x + b)/((c*x^2 + b*x + a)*(e*x + d)^(5/2)),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x+b)/(e*x+d)**(5/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

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GIAC/XCAS [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x + b)/((c*x^2 + b*x + a)*(e*x + d)^(5/2)),x, algorithm="giac")

[Out]

Timed out