3.367 \(\int \frac{1}{(d+e x)^{5/2} (b x+c x^2)} \, dx\)
Optimal. Leaf size=138 \[ \frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{5/2}}-\frac{2 e (2 c d-b e)}{d^2 \sqrt{d+e x} (c d-b e)^2}-\frac{2 e}{3 d (d+e x)^{3/2} (c d-b e)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{5/2}} \]
[Out]
(-2*e)/(3*d*(c*d - b*e)*(d + e*x)^(3/2)) - (2*e*(2*c*d - b*e))/(d^2*(c*d - b*e)^2*Sqrt[d + e*x]) - (2*ArcTanh[
Sqrt[d + e*x]/Sqrt[d]])/(b*d^(5/2)) + (2*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*(c*d - b
*e)^(5/2))
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Rubi [A] time = 0.260793, antiderivative size = 138, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used =
{709, 828, 826, 1166, 208} \[ \frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{5/2}}-\frac{2 e (2 c d-b e)}{d^2 \sqrt{d+e x} (c d-b e)^2}-\frac{2 e}{3 d (d+e x)^{3/2} (c d-b e)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/((d + e*x)^(5/2)*(b*x + c*x^2)),x]
[Out]
(-2*e)/(3*d*(c*d - b*e)*(d + e*x)^(3/2)) - (2*e*(2*c*d - b*e))/(d^2*(c*d - b*e)^2*Sqrt[d + e*x]) - (2*ArcTanh[
Sqrt[d + e*x]/Sqrt[d]])/(b*d^(5/2)) + (2*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*(c*d - b
*e)^(5/2))
Rule 709
Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d - b*e - c
*e*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[m, -1]
Rule 828
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), 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] && FractionQ[m] && LtQ[m, -1]
Rule 826
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
Rule 1166
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]
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]
Rubi steps
\begin{align*} \int \frac{1}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx &=-\frac{2 e}{3 d (c d-b e) (d+e x)^{3/2}}+\frac{\int \frac{c d-b e-c e x}{(d+e x)^{3/2} \left (b x+c x^2\right )} \, dx}{d (c d-b e)}\\ &=-\frac{2 e}{3 d (c d-b e) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{d^2 (c d-b e)^2 \sqrt{d+e x}}+\frac{\int \frac{(c d-b e)^2-c e (2 c d-b e) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{d^2 (c d-b e)^2}\\ &=-\frac{2 e}{3 d (c d-b e) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{d^2 (c d-b e)^2 \sqrt{d+e x}}+\frac{2 \operatorname{Subst}\left (\int \frac{e (c d-b e)^2+c d e (2 c d-b e)-c e (2 c d-b e) x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{d^2 (c d-b e)^2}\\ &=-\frac{2 e}{3 d (c d-b e) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{d^2 (c d-b e)^2 \sqrt{d+e x}}+\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b d^2}-\frac{\left (2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b (c d-b e)^2}\\ &=-\frac{2 e}{3 d (c d-b e) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{d^2 (c d-b e)^2 \sqrt{d+e x}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{5/2}}+\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.0267906, size = 83, normalized size = 0.6 \[ -\frac{2 \left (c d \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c (d+e x)}{c d-b e}\right )+(b e-c d) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{e x}{d}+1\right )\right )}{3 b d (d+e x)^{3/2} (c d-b e)} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((d + e*x)^(5/2)*(b*x + c*x^2)),x]
[Out]
(-2*(c*d*Hypergeometric2F1[-3/2, 1, -1/2, (c*(d + e*x))/(c*d - b*e)] + (-(c*d) + b*e)*Hypergeometric2F1[-3/2,
1, -1/2, 1 + (e*x)/d]))/(3*b*d*(c*d - b*e)*(d + e*x)^(3/2))
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Maple [A] time = 0.224, size = 147, normalized size = 1.1 \begin{align*} 2\,{\frac{b{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}-4\,{\frac{ce}{d \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}+{\frac{2\,e}{3\,d \left ( be-cd \right ) } \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{c}^{3}}{ \left ( be-cd \right ) ^{2}b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{1}{b{d}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x+d)^(5/2)/(c*x^2+b*x),x)
[Out]
2/d^2/(b*e-c*d)^2/(e*x+d)^(1/2)*b*e^2-4*e/d/(b*e-c*d)^2/(e*x+d)^(1/2)*c+2/3*e/d/(b*e-c*d)/(e*x+d)^(3/2)-2/(b*e
-c*d)^2*c^3/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))-2*arctanh((e*x+d)^(1/2)/d^(1/2))
/b/d^(5/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.59143, size = 3040, normalized size = 22.03 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="fricas")
[Out]
[1/3*(3*(c^2*d^3*e^2*x^2 + 2*c^2*d^4*e*x + c^2*d^5)*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)) + 3*(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + (c^2*d^2*e^2 - 2*b
*c*d*e^3 + b^2*e^4)*x^2 + 2*(c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt
(d) + 2*d)/x) - 2*(7*b*c*d^3*e - 4*b^2*d^2*e^2 + 3*(2*b*c*d^2*e^2 - b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c^2*d^7 -
2*b^2*c*d^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e - 2*b^2*c*d
^5*e^2 + b^3*d^4*e^3)*x), 1/3*(6*(c^2*d^3*e^2*x^2 + 2*c^2*d^4*e*x + c^2*d^5)*sqrt(-c/(c*d - b*e))*arctan(-(c*d
- b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + 3*(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + (c^2*d^2*
e^2 - 2*b*c*d*e^3 + b^2*e^4)*x^2 + 2*(c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x
+ d)*sqrt(d) + 2*d)/x) - 2*(7*b*c*d^3*e - 4*b^2*d^2*e^2 + 3*(2*b*c*d^2*e^2 - b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c
^2*d^7 - 2*b^2*c*d^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e -
2*b^2*c*d^5*e^2 + b^3*d^4*e^3)*x), 1/3*(6*(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + (c^2*d^2*e^2 - 2*b*c*d*e^3 +
b^2*e^4)*x^2 + 2*(c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 3*(c^2
*d^3*e^2*x^2 + 2*c^2*d^4*e*x + c^2*d^5)*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)) - 2*(7*b*c*d^3*e - 4*b^2*d^2*e^2 + 3*(2*b*c*d^2*e^2 - b^2*d*e^3)*x)*sqrt(
e*x + d))/(b*c^2*d^7 - 2*b^2*c*d^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(
b*c^2*d^6*e - 2*b^2*c*d^5*e^2 + b^3*d^4*e^3)*x), 2/3*(3*(c^2*d^3*e^2*x^2 + 2*c^2*d^4*e*x + c^2*d^5)*sqrt(-c/(c
*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + 3*(c^2*d^4 - 2*b*c*d^3*e +
b^2*d^2*e^2 + (c^2*d^2*e^2 - 2*b*c*d*e^3 + b^2*e^4)*x^2 + 2*(c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(-d
)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (7*b*c*d^3*e - 4*b^2*d^2*e^2 + 3*(2*b*c*d^2*e^2 - b^2*d*e^3)*x)*sqrt(e*x
+ d))/(b*c^2*d^7 - 2*b^2*c*d^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^
2*d^6*e - 2*b^2*c*d^5*e^2 + b^3*d^4*e^3)*x)]
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Sympy [A] time = 21.7701, size = 133, normalized size = 0.96 \begin{align*} \frac{2 e}{3 d \left (d + e x\right )^{\frac{3}{2}} \left (b e - c d\right )} + \frac{2 e \left (b e - 2 c d\right )}{d^{2} \sqrt{d + e x} \left (b e - c d\right )^{2}} - \frac{2 c^{2} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b \sqrt{\frac{b e - c d}{c}} \left (b e - c d\right )^{2}} + \frac{2 \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b d^{2} \sqrt{- d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)**(5/2)/(c*x**2+b*x),x)
[Out]
2*e/(3*d*(d + e*x)**(3/2)*(b*e - c*d)) + 2*e*(b*e - 2*c*d)/(d**2*sqrt(d + e*x)*(b*e - c*d)**2) - 2*c**2*atan(s
qrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*sqrt((b*e - c*d)/c)*(b*e - c*d)**2) + 2*atan(sqrt(d + e*x)/sqrt(-d))/(b*d
**2*sqrt(-d))
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Giac [A] time = 1.32073, size = 235, normalized size = 1.7 \begin{align*} -\frac{2 \, c^{3} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} \sqrt{-c^{2} d + b c e}} - \frac{2 \,{\left (6 \,{\left (x e + d\right )} c d e + c d^{2} e - 3 \,{\left (x e + d\right )} b e^{2} - b d e^{2}\right )}}{3 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}{\left (x e + d\right )}^{\frac{3}{2}}} + \frac{2 \, \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d} d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="giac")
[Out]
-2*c^3*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*sqrt(-c^2*d + b*c*e))
- 2/3*(6*(x*e + d)*c*d*e + c*d^2*e - 3*(x*e + d)*b*e^2 - b*d*e^2)/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*(x*e
+ d)^(3/2)) + 2*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)*d^2)