3.1236 \(\int \frac{A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx\)

Optimal. Leaf size=225 \[ \frac{2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 \sqrt{d+e x} (c d-b e)^3}-\frac{2 c^{5/2} (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{7/2}}+\frac{2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d^2 (d+e x)^{3/2} (c d-b e)^2}+\frac{2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{7/2}} \]

[Out]

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

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Rubi [A]  time = 1.12145, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231 \[ \frac{2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 \sqrt{d+e x} (c d-b e)^3}-\frac{2 c^{5/2} (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{7/2}}+\frac{2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d^2 (d+e x)^{3/2} (c d-b e)^2}+\frac{2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{7/2}} \]

Antiderivative was successfully verified.

[In]  Int[(A + B*x)/((d + e*x)^(7/2)*(b*x + c*x^2)),x]

[Out]

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

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Rubi in Sympy [A]  time = 110.372, size = 212, normalized size = 0.94 \[ - \frac{2 A \operatorname{atanh}{\left (\frac{\sqrt{d + e x}}{\sqrt{d}} \right )}}{b d^{\frac{7}{2}}} + \frac{2 \left (A e - B d\right )}{5 d \left (d + e x\right )^{\frac{5}{2}} \left (b e - c d\right )} + \frac{2 \left (A b e^{2} - 2 A c d e + B c d^{2}\right )}{3 d^{2} \left (d + e x\right )^{\frac{3}{2}} \left (b e - c d\right )^{2}} + \frac{2 \left (A e \left (b e - c d\right )^{2} - c d \left (A b e^{2} - 2 A c d e + B c d^{2}\right )\right )}{d^{3} \sqrt{d + e x} \left (b e - c d\right )^{3}} + \frac{2 c^{\frac{5}{2}} \left (A c - B b\right ) \operatorname{atan}{\left (\frac{\sqrt{c} \sqrt{d + e x}}{\sqrt{b e - c d}} \right )}}{b \left (b e - c d\right )^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)/(e*x+d)**(7/2)/(c*x**2+b*x),x)

[Out]

-2*A*atanh(sqrt(d + e*x)/sqrt(d))/(b*d**(7/2)) + 2*(A*e - B*d)/(5*d*(d + e*x)**(
5/2)*(b*e - c*d)) + 2*(A*b*e**2 - 2*A*c*d*e + B*c*d**2)/(3*d**2*(d + e*x)**(3/2)
*(b*e - c*d)**2) + 2*(A*e*(b*e - c*d)**2 - c*d*(A*b*e**2 - 2*A*c*d*e + B*c*d**2)
)/(d**3*sqrt(d + e*x)*(b*e - c*d)**3) + 2*c**(5/2)*(A*c - B*b)*atan(sqrt(c)*sqrt
(d + e*x)/sqrt(b*e - c*d))/(b*(b*e - c*d)**(7/2))

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Mathematica [A]  time = 1.40276, size = 223, normalized size = 0.99 \[ \frac{2 B c^2 d^3-2 A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{d^3 \sqrt{d+e x} (c d-b e)^3}+\frac{2 c^{5/2} (A c-b B) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{7/2}}+\frac{2 \left (A e (b e-2 c d)+B c d^2\right )}{3 d^2 (d+e x)^{3/2} (c d-b e)^2}+\frac{2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{7/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(A + B*x)/((d + e*x)^(7/2)*(b*x + c*x^2)),x]

[Out]

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

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Maple [A]  time = 0.033, size = 350, normalized size = 1.6 \[{\frac{2\,Ae}{5\,d \left ( be-cd \right ) } \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,B}{5\,be-5\,cd} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,Ab{e}^{2}}{3\,{d}^{2} \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,Ace}{3\,d \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Bc}{3\, \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{A{b}^{2}{e}^{3}}{{d}^{3} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}-6\,{\frac{Abc{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}+6\,{\frac{A{c}^{2}e}{d \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}-2\,{\frac{B{c}^{2}}{ \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}+2\,{\frac{{c}^{4}A}{ \left ( be-cd \right ) ^{3}b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{B{c}^{3}}{ \left ( be-cd \right ) ^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{A}{b{d}^{7/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x+A)/(e*x+d)^(7/2)/(c*x^2+b*x),x)

[Out]

2/5/d/(b*e-c*d)/(e*x+d)^(5/2)*A*e-2/5/(b*e-c*d)/(e*x+d)^(5/2)*B+2/3/d^2/(b*e-c*d
)^2/(e*x+d)^(3/2)*A*b*e^2-4/3/d/(b*e-c*d)^2/(e*x+d)^(3/2)*A*c*e+2/3/(b*e-c*d)^2/
(e*x+d)^(3/2)*B*c+2/d^3/(b*e-c*d)^3/(e*x+d)^(1/2)*A*b^2*e^3-6/d^2/(b*e-c*d)^3/(e
*x+d)^(1/2)*A*b*c*e^2+6/d/(b*e-c*d)^3/(e*x+d)^(1/2)*A*c^2*e-2/(b*e-c*d)^3/(e*x+d
)^(1/2)*B*c^2+2/(b*e-c*d)^3*c^4/b/((b*e-c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b
*e-c*d)*c)^(1/2))*A-2/(b*e-c*d)^3*c^3/((b*e-c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)
/((b*e-c*d)*c)^(1/2))*B-2*A*arctanh((e*x+d)^(1/2)/d^(1/2))/b/d^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/((c*x^2 + b*x)*(e*x + d)^(7/2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.18062, size = 1, normalized size = 0. \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/((c*x^2 + b*x)*(e*x + d)^(7/2)),x, algorithm="fricas")

[Out]

[1/15*(15*((B*b*c^2 - A*c^3)*d^3*e^2*x^2 + 2*(B*b*c^2 - A*c^3)*d^4*e*x + (B*b*c^
2 - A*c^3)*d^5)*sqrt(e*x + d)*sqrt(d)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b
*e - 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) + 15*(A*c^3*d^5
- 3*A*b*c^2*d^4*e + 3*A*b^2*c*d^3*e^2 - A*b^3*d^2*e^3 + (A*c^3*d^3*e^2 - 3*A*b*
c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 + 2*(A*c^3*d^4*e - 3*A*b*c^2*d^3*
e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x)*sqrt(e*x + d)*log(((e*x + 2*d)*sqrt(d)
- 2*sqrt(e*x + d)*d)/x) + 2*(23*B*b*c^2*d^5 - 23*A*b^3*d^2*e^3 - (11*B*b^2*c +
58*A*b*c^2)*d^4*e + 3*(B*b^3 + 22*A*b^2*c)*d^3*e^2 + 15*(B*b*c^2*d^3*e^2 - 3*A*b
*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 + 5*(7*B*b*c^2*d^4*e + 21*A*b^2*
c*d^2*e^3 - 7*A*b^3*d*e^4 - (B*b^2*c + 20*A*b*c^2)*d^3*e^2)*x)*sqrt(d))/((b*c^3*
d^8 - 3*b^2*c^2*d^7*e + 3*b^3*c*d^6*e^2 - b^4*d^5*e^3 + (b*c^3*d^6*e^2 - 3*b^2*c
^2*d^5*e^3 + 3*b^3*c*d^4*e^4 - b^4*d^3*e^5)*x^2 + 2*(b*c^3*d^7*e - 3*b^2*c^2*d^6
*e^2 + 3*b^3*c*d^5*e^3 - b^4*d^4*e^4)*x)*sqrt(e*x + d)*sqrt(d)), -1/15*(30*((B*b
*c^2 - A*c^3)*d^3*e^2*x^2 + 2*(B*b*c^2 - A*c^3)*d^4*e*x + (B*b*c^2 - A*c^3)*d^5)
*sqrt(e*x + d)*sqrt(d)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(-c/(c*d - b
*e))/(sqrt(e*x + d)*c)) - 15*(A*c^3*d^5 - 3*A*b*c^2*d^4*e + 3*A*b^2*c*d^3*e^2 -
A*b^3*d^2*e^3 + (A*c^3*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5
)*x^2 + 2*(A*c^3*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x)
*sqrt(e*x + d)*log(((e*x + 2*d)*sqrt(d) - 2*sqrt(e*x + d)*d)/x) - 2*(23*B*b*c^2*
d^5 - 23*A*b^3*d^2*e^3 - (11*B*b^2*c + 58*A*b*c^2)*d^4*e + 3*(B*b^3 + 22*A*b^2*c
)*d^3*e^2 + 15*(B*b*c^2*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^
5)*x^2 + 5*(7*B*b*c^2*d^4*e + 21*A*b^2*c*d^2*e^3 - 7*A*b^3*d*e^4 - (B*b^2*c + 20
*A*b*c^2)*d^3*e^2)*x)*sqrt(d))/((b*c^3*d^8 - 3*b^2*c^2*d^7*e + 3*b^3*c*d^6*e^2 -
b^4*d^5*e^3 + (b*c^3*d^6*e^2 - 3*b^2*c^2*d^5*e^3 + 3*b^3*c*d^4*e^4 - b^4*d^3*e^
5)*x^2 + 2*(b*c^3*d^7*e - 3*b^2*c^2*d^6*e^2 + 3*b^3*c*d^5*e^3 - b^4*d^4*e^4)*x)*
sqrt(e*x + d)*sqrt(d)), 1/15*(15*((B*b*c^2 - A*c^3)*d^3*e^2*x^2 + 2*(B*b*c^2 - A
*c^3)*d^4*e*x + (B*b*c^2 - A*c^3)*d^5)*sqrt(e*x + d)*sqrt(-d)*sqrt(c/(c*d - b*e)
)*log((c*e*x + 2*c*d - b*e - 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c
*x + b)) + 30*(A*c^3*d^5 - 3*A*b*c^2*d^4*e + 3*A*b^2*c*d^3*e^2 - A*b^3*d^2*e^3 +
(A*c^3*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 + 2*(A*c^
3*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x)*sqrt(e*x + d)*
arctan(d/(sqrt(e*x + d)*sqrt(-d))) + 2*(23*B*b*c^2*d^5 - 23*A*b^3*d^2*e^3 - (11*
B*b^2*c + 58*A*b*c^2)*d^4*e + 3*(B*b^3 + 22*A*b^2*c)*d^3*e^2 + 15*(B*b*c^2*d^3*e
^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 + 5*(7*B*b*c^2*d^4*e +
21*A*b^2*c*d^2*e^3 - 7*A*b^3*d*e^4 - (B*b^2*c + 20*A*b*c^2)*d^3*e^2)*x)*sqrt(-d
))/((b*c^3*d^8 - 3*b^2*c^2*d^7*e + 3*b^3*c*d^6*e^2 - b^4*d^5*e^3 + (b*c^3*d^6*e^
2 - 3*b^2*c^2*d^5*e^3 + 3*b^3*c*d^4*e^4 - b^4*d^3*e^5)*x^2 + 2*(b*c^3*d^7*e - 3*
b^2*c^2*d^6*e^2 + 3*b^3*c*d^5*e^3 - b^4*d^4*e^4)*x)*sqrt(e*x + d)*sqrt(-d)), -2/
15*(15*((B*b*c^2 - A*c^3)*d^3*e^2*x^2 + 2*(B*b*c^2 - A*c^3)*d^4*e*x + (B*b*c^2 -
A*c^3)*d^5)*sqrt(e*x + d)*sqrt(-d)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqr
t(-c/(c*d - b*e))/(sqrt(e*x + d)*c)) - 15*(A*c^3*d^5 - 3*A*b*c^2*d^4*e + 3*A*b^2
*c*d^3*e^2 - A*b^3*d^2*e^3 + (A*c^3*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^
4 - A*b^3*e^5)*x^2 + 2*(A*c^3*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*
b^3*d*e^4)*x)*sqrt(e*x + d)*arctan(d/(sqrt(e*x + d)*sqrt(-d))) - (23*B*b*c^2*d^5
- 23*A*b^3*d^2*e^3 - (11*B*b^2*c + 58*A*b*c^2)*d^4*e + 3*(B*b^3 + 22*A*b^2*c)*d
^3*e^2 + 15*(B*b*c^2*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*
x^2 + 5*(7*B*b*c^2*d^4*e + 21*A*b^2*c*d^2*e^3 - 7*A*b^3*d*e^4 - (B*b^2*c + 20*A*
b*c^2)*d^3*e^2)*x)*sqrt(-d))/((b*c^3*d^8 - 3*b^2*c^2*d^7*e + 3*b^3*c*d^6*e^2 - b
^4*d^5*e^3 + (b*c^3*d^6*e^2 - 3*b^2*c^2*d^5*e^3 + 3*b^3*c*d^4*e^4 - b^4*d^3*e^5)
*x^2 + 2*(b*c^3*d^7*e - 3*b^2*c^2*d^6*e^2 + 3*b^3*c*d^5*e^3 - b^4*d^4*e^4)*x)*sq
rt(e*x + d)*sqrt(-d))]

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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((B*x+A)/(e*x+d)**(7/2)/(c*x**2+b*x),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.307585, size = 521, normalized size = 2.32 \[ \frac{2 \,{\left (B b c^{3} - A c^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} \sqrt{-c^{2} d + b c e}} + \frac{2 \,{\left (15 \,{\left (x e + d\right )}^{2} B c^{2} d^{3} + 5 \,{\left (x e + d\right )} B c^{2} d^{4} + 3 \, B c^{2} d^{5} - 45 \,{\left (x e + d\right )}^{2} A c^{2} d^{2} e - 5 \,{\left (x e + d\right )} B b c d^{3} e - 10 \,{\left (x e + d\right )} A c^{2} d^{3} e - 6 \, B b c d^{4} e - 3 \, A c^{2} d^{4} e + 45 \,{\left (x e + d\right )}^{2} A b c d e^{2} + 15 \,{\left (x e + d\right )} A b c d^{2} e^{2} + 3 \, B b^{2} d^{3} e^{2} + 6 \, A b c d^{3} e^{2} - 15 \,{\left (x e + d\right )}^{2} A b^{2} e^{3} - 5 \,{\left (x e + d\right )} A b^{2} d e^{3} - 3 \, A b^{2} d^{2} e^{3}\right )}}{15 \,{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} + \frac{2 \, A \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/((c*x^2 + b*x)*(e*x + d)^(7/2)),x, algorithm="giac")

[Out]

2*(B*b*c^3 - A*c^4)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b*c^3*d^3 - 3
*b^2*c^2*d^2*e + 3*b^3*c*d*e^2 - b^4*e^3)*sqrt(-c^2*d + b*c*e)) + 2/15*(15*(x*e
+ d)^2*B*c^2*d^3 + 5*(x*e + d)*B*c^2*d^4 + 3*B*c^2*d^5 - 45*(x*e + d)^2*A*c^2*d^
2*e - 5*(x*e + d)*B*b*c*d^3*e - 10*(x*e + d)*A*c^2*d^3*e - 6*B*b*c*d^4*e - 3*A*c
^2*d^4*e + 45*(x*e + d)^2*A*b*c*d*e^2 + 15*(x*e + d)*A*b*c*d^2*e^2 + 3*B*b^2*d^3
*e^2 + 6*A*b*c*d^3*e^2 - 15*(x*e + d)^2*A*b^2*e^3 - 5*(x*e + d)*A*b^2*d*e^3 - 3*
A*b^2*d^2*e^3)/((c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*(x*e +
d)^(5/2)) + 2*A*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)*d^3)