3.371 \(\int \frac{(d+e x)^{5/2}}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=159 \[ -\frac{(c d-b e)^{3/2} (b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{3/2}}+\frac{d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac{e \sqrt{d+e x} (2 c d-b e)}{b^2 c} \]

[Out]

(e*(2*c*d - b*e)*Sqrt[d + e*x])/(b^2*c) - ((d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(b^2*(b*x + c*x^2)) + (d^(
3/2)*(4*c*d - 5*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b^3 - ((c*d - b*e)^(3/2)*(4*c*d + b*e)*ArcTanh[(Sqrt[c]*S
qrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*c^(3/2))

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Rubi [A]  time = 0.3019, antiderivative size = 159, 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 = {738, 824, 826, 1166, 208} \[ -\frac{(c d-b e)^{3/2} (b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{3/2}}+\frac{d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac{e \sqrt{d+e x} (2 c d-b e)}{b^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(2*c*d - b*e)*Sqrt[d + e*x])/(b^2*c) - ((d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(b^2*(b*x + c*x^2)) + (d^(
3/2)*(4*c*d - 5*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b^3 - ((c*d - b*e)^(3/2)*(4*c*d + b*e)*ArcTanh[(Sqrt[c]*S
qrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*c^(3/2))

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), 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] && FractionQ[m] && GtQ[m, 0]

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{(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx &=-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{\sqrt{d+e x} \left (\frac{1}{2} d (4 c d-5 b e)-\frac{1}{2} e (2 c d-b e) x\right )}{b x+c x^2} \, dx}{b^2}\\ &=\frac{e (2 c d-b e) \sqrt{d+e x}}{b^2 c}-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{\frac{1}{2} c d^2 (4 c d-5 b e)+\frac{1}{2} e \left (2 c^2 d^2-2 b c d e-b^2 e^2\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{b^2 c}\\ &=\frac{e (2 c d-b e) \sqrt{d+e x}}{b^2 c}-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} c d^2 e (4 c d-5 b e)-\frac{1}{2} d e \left (2 c^2 d^2-2 b c d e-b^2 e^2\right )+\frac{1}{2} e \left (2 c^2 d^2-2 b c d e-b^2 e^2\right ) 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 )}{b^2 c}\\ &=\frac{e (2 c d-b e) \sqrt{d+e x}}{b^2 c}-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\left (c d^2 (4 c d-5 b e)\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^3}+\frac{\left ((c d-b e)^2 (4 c d+b e)\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^3 c}\\ &=\frac{e (2 c d-b e) \sqrt{d+e x}}{b^2 c}-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}+\frac{d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(c d-b e)^{3/2} (4 c d+b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.275935, size = 159, normalized size = 1. \[ \frac{-\frac{b \sqrt{d+e x} \left (b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )}{c x (b+c x)}-\frac{\sqrt{c d-b e} \left (-b^2 e^2-3 b c d e+4 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{c^{3/2}}+d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*Sqrt[d + e*x]*(2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x)))/(c*x*(b + c*x))) + d^(3/2)*(4*c*d - 5*b*e)*
ArcTanh[Sqrt[d + e*x]/Sqrt[d]] - (Sqrt[c*d - b*e]*(4*c^2*d^2 - 3*b*c*d*e - b^2*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d +
e*x])/Sqrt[c*d - b*e]])/c^(3/2))/b^3

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Maple [B]  time = 0.242, size = 313, normalized size = 2. \begin{align*} -{\frac{{e}^{3}}{c \left ( cex+be \right ) }\sqrt{ex+d}}+2\,{\frac{{e}^{2}\sqrt{ex+d}d}{b \left ( cex+be \right ) }}-{\frac{ce{d}^{2}}{{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}+{\frac{{e}^{3}}{c}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}}+2\,{\frac{d{e}^{2}}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-7\,{\frac{ce{d}^{2}}{{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{c}^{2}{d}^{3}}{{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{{d}^{2}}{{b}^{2}x}\sqrt{ex+d}}-5\,{\frac{e{d}^{3/2}}{{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{{d}^{5/2}c}{{b}^{3}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((e*x+d)^(5/2)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.0379, size = 2122, normalized size = 13.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt((c*d - b*e)/
c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x
^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c*d^2 + (2*b
*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/(b^3*c^2*x^2 + b^4*c*x), -1/2*(2*((4*c^3*d^2 - 3*b*c^2*d*e
- b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt
(-(c*d - b*e)/c)/(c*d - b*e)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(d)*log((e
*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d)
)/(b^3*c^2*x^2 + b^4*c*x), -1/2*(2*((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(-d)*ar
ctan(sqrt(e*x + d)*sqrt(-d)/d) + ((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3
*e^2)*x)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*
(b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/(b^3*c^2*x^2 + b^4*c*x), -(((4*c^3*d^2 -
3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x
+ d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqr
t(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/
(b^3*c^2*x^2 + b^4*c*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.38802, size = 369, normalized size = 2.32 \begin{align*} -\frac{{\left (4 \, c d^{3} - 5 \, b d^{2} e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d}} + \frac{{\left (4 \, c^{3} d^{3} - 7 \, b c^{2} d^{2} e + 2 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b^{3} c} - \frac{2 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{2} d^{2} e - 2 \, \sqrt{x e + d} c^{2} d^{3} e - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} b c d e^{2} + 3 \, \sqrt{x e + d} b c d^{2} e^{2} +{\left (x e + d\right )}^{\frac{3}{2}} b^{2} e^{3} - \sqrt{x e + d} b^{2} d e^{3}}{{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )} b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*c*d^3 - 5*b*d^2*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)) + (4*c^3*d^3 - 7*b*c^2*d^2*e + 2*b^2*c*d*
e^2 + b^3*e^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b^3*c) - (2*(x*e + d)^(3/2)*
c^2*d^2*e - 2*sqrt(x*e + d)*c^2*d^3*e - 2*(x*e + d)^(3/2)*b*c*d*e^2 + 3*sqrt(x*e + d)*b*c*d^2*e^2 + (x*e + d)^
(3/2)*b^2*e^3 - sqrt(x*e + d)*b^2*d*e^3)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)*b^
2*c)