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

Optimal. Leaf size=118 \[ \frac{2 e \sqrt{d+e x} (2 c d-b e)}{c^2}+\frac{2 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{5/2}}-\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 e (d+e x)^{3/2}}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.227992, antiderivative size = 118, 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 = {703, 824, 826, 1166, 208} \[ \frac{2 e \sqrt{d+e x} (2 c d-b e)}{c^2}+\frac{2 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{5/2}}-\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 e (d+e x)^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/(b*x + c*x^2),x]

[Out]

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

Rule 703

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

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

Mathematica [A]  time = 0.123229, size = 107, normalized size = 0.91 \[ \frac{2 e \sqrt{d+e x} (-3 b e+7 c d+c e x)}{3 c^2}+\frac{2 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{5/2}}-\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.223, size = 237, normalized size = 2. \begin{align*}{\frac{2\,e}{3\,c} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-2\,{\frac{\sqrt{ex+d}b{e}^{2}}{{c}^{2}}}+4\,{\frac{de\sqrt{ex+d}}{c}}+2\,{\frac{{e}^{3}{b}^{2}}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-6\,{\frac{bd{e}^{2}}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+6\,{\frac{e{d}^{2}}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{c{d}^{3}}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{{d}^{5/2}}{b}{\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),x)

[Out]

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

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.86695, size = 1350, normalized size = 11.44 \begin{align*} \left [\frac{3 \, c^{2} d^{\frac{5}{2}} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) + 3 \,{\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt{\frac{c d - b e}{c}} \log \left (\frac{c e x + 2 \, c d - b e + 2 \, \sqrt{e x + d} c \sqrt{\frac{c d - b e}{c}}}{c x + b}\right ) + 2 \,{\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt{e x + d}}{3 \, b c^{2}}, \frac{3 \, c^{2} d^{\frac{5}{2}} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) + 6 \,{\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt{-\frac{c d - b e}{c}} \arctan \left (-\frac{\sqrt{e x + d} c \sqrt{-\frac{c d - b e}{c}}}{c d - b e}\right ) + 2 \,{\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt{e x + d}}{3 \, b c^{2}}, \frac{6 \, c^{2} \sqrt{-d} d^{2} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) + 3 \,{\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt{\frac{c d - b e}{c}} \log \left (\frac{c e x + 2 \, c d - b e + 2 \, \sqrt{e x + d} c \sqrt{\frac{c d - b e}{c}}}{c x + b}\right ) + 2 \,{\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt{e x + d}}{3 \, b c^{2}}, \frac{2 \,{\left (3 \, c^{2} \sqrt{-d} d^{2} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) + 3 \,{\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt{-\frac{c d - b e}{c}} \arctan \left (-\frac{\sqrt{e x + d} c \sqrt{-\frac{c d - b e}{c}}}{c d - b e}\right ) +{\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt{e x + d}\right )}}{3 \, b c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*c^2*d^(5/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 3*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*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*c*e^2*x + 7*b*c
*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2), 1/3*(3*c^2*d^(5/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 6*
(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e))
+ 2*(b*c*e^2*x + 7*b*c*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2), 1/3*(6*c^2*sqrt(-d)*d^2*arctan(sqrt(e*x + d)*
sqrt(-d)/d) + 3*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*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*c*e^2*x + 7*b*c*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2), 2/3*(3*c^2*
sqrt(-d)*d^2*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 3*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*sqrt(-(c*d - b*e)/c)*arctan(
-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + (b*c*e^2*x + 7*b*c*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2
)]

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Sympy [A]  time = 56.1243, size = 119, normalized size = 1.01 \begin{align*} \frac{2 e \left (d + e x\right )^{\frac{3}{2}}}{3 c} + \frac{\sqrt{d + e x} \left (- 2 b e^{2} + 4 c d e\right )}{c^{2}} + \frac{2 d^{3} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b \sqrt{- d}} + \frac{2 \left (b e - c d\right )^{3} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b c^{3} \sqrt{\frac{b e - c d}{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),x)

[Out]

2*e*(d + e*x)**(3/2)/(3*c) + sqrt(d + e*x)*(-2*b*e**2 + 4*c*d*e)/c**2 + 2*d**3*atan(sqrt(d + e*x)/sqrt(-d))/(b
*sqrt(-d)) + 2*(b*e - c*d)**3*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**3*sqrt((b*e - c*d)/c))

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Giac [A]  time = 1.34448, size = 217, normalized size = 1.84 \begin{align*} \frac{2 \, d^{3} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d}} - \frac{2 \,{\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, 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 c^{2}} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} c^{2} e + 6 \, \sqrt{x e + d} c^{2} d e - 3 \, \sqrt{x e + d} b c e^{2}\right )}}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^3*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) - 2*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*arct
an(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c^2) + 2/3*((x*e + d)^(3/2)*c^2*e + 6*sqrt(x*
e + d)*c^2*d*e - 3*sqrt(x*e + d)*b*c*e^2)/c^3