3.363 \(\int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx\)

Optimal. Leaf size=92 \[ \frac {2 (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}}-\frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 e \sqrt {d+e x}}{c} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {703, 826, 1166, 208} \[ \frac {2 (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}}-\frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 e \sqrt {d+e x}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)/(b*x + c*x^2),x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 97, normalized size = 1.05 \[ \frac {2 \left (b \sqrt {c} e \sqrt {d+e x}+(c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )-c^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{b c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)/(b*x + c*x^2),x]

[Out]

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

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fricas [A]  time = 1.18, size = 447, normalized size = 4.86 \[ \left [\frac {c d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} b e - {\left (c d - b e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {c d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} b e + 2 \, {\left (c d - b e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )}{b c}, \frac {2 \, c \sqrt {-d} d \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + 2 \, \sqrt {e x + d} b e - {\left (c d - b e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {2 \, {\left (c \sqrt {-d} d \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + \sqrt {e x + d} b e + {\left (c d - b e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )\right )}}{b c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 112, normalized size = 1.22 \[ \frac {2 \, d^{2} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, \sqrt {x e + d} e}{c} - \frac {2 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

2*d^2*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) + 2*sqrt(x*e + d)*e/c - 2*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*ar
ctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c)

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maple [B]  time = 0.05, size = 159, normalized size = 1.73 \[ -\frac {2 b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}-\frac {2 c \,d^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}+\frac {4 d e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 d^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \sqrt {e x +d}\, e}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)/(c*x^2+b*x),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive or negative?

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mupad [B]  time = 0.37, size = 697, normalized size = 7.58 \[ \frac {2\,e\,\sqrt {d+e\,x}}{c}-\frac {2\,\mathrm {atanh}\left (\frac {16\,b^3\,e^6\,\sqrt {d^3}\,\sqrt {d+e\,x}}{16\,b^3\,d^2\,e^6-64\,b^2\,c\,d^3\,e^5+96\,b\,c^2\,d^4\,e^4-48\,c^3\,d^5\,e^3}+\frac {48\,c^2\,d^3\,e^3\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}+\frac {64\,b^2\,d\,e^5\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}-\frac {96\,b\,c\,d^2\,e^4\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}\right )\,\sqrt {d^3}}{b}+\frac {2\,\mathrm {atanh}\left (\frac {48\,d^3\,e^3\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{48\,c^3\,d^5\,e^3-80\,b^3\,d^2\,e^6-144\,b\,c^2\,d^4\,e^4+160\,b^2\,c\,d^3\,e^5+\frac {16\,b^4\,d\,e^7}{c}}+\frac {16\,b^2\,d\,e^5\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{16\,b^4\,c\,d\,e^7-80\,b^3\,c^2\,d^2\,e^6+160\,b^2\,c^3\,d^3\,e^5-144\,b\,c^4\,d^4\,e^4+48\,c^5\,d^5\,e^3}-\frac {48\,b\,d^2\,e^4\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{16\,b^4\,d\,e^7-80\,b^3\,c\,d^2\,e^6+160\,b^2\,c^2\,d^3\,e^5-144\,b\,c^3\,d^4\,e^4+48\,c^4\,d^5\,e^3}\right )\,\sqrt {-c^3\,{\left (b\,e-c\,d\right )}^3}}{b\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(3/2)/(b*x + c*x^2),x)

[Out]

(2*e*(d + e*x)^(1/2))/c - (2*atanh((16*b^3*e^6*(d^3)^(1/2)*(d + e*x)^(1/2))/(16*b^3*d^2*e^6 - 48*c^3*d^5*e^3 +
96*b*c^2*d^4*e^4 - 64*b^2*c*d^3*e^5) + (48*c^2*d^3*e^3*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d^3*e^5 + 48*c^2*
d^5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4) + (64*b^2*d*e^5*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d^3*e^5 +
48*c^2*d^5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4) - (96*b*c*d^2*e^4*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d
^3*e^5 + 48*c^2*d^5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4))*(d^3)^(1/2))/b + (2*atanh((48*d^3*e^3*(d + e*x
)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(48*c^3*d^5*e^3 - 80*b^3*d^2*e^6 - 14
4*b*c^2*d^4*e^4 + 160*b^2*c*d^3*e^5 + (16*b^4*d*e^7)/c) + (16*b^2*d*e^5*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3
+ 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(48*c^5*d^5*e^3 - 144*b*c^4*d^4*e^4 + 160*b^2*c^3*d^3*e^5 - 80*b^3*
c^2*d^2*e^6 + 16*b^4*c*d*e^7) - (48*b*d^2*e^4*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c
^5*d^2*e)^(1/2))/(16*b^4*d*e^7 + 48*c^4*d^5*e^3 - 144*b*c^3*d^4*e^4 - 80*b^3*c*d^2*e^6 + 160*b^2*c^2*d^3*e^5))
*(-c^3*(b*e - c*d)^3)^(1/2))/(b*c^3)

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sympy [A]  time = 33.67, size = 92, normalized size = 1.00 \[ \frac {2 e \sqrt {d + e x}}{c} + \frac {2 d^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} - \frac {2 \left (b e - c d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{2} \sqrt {\frac {b e - c d}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)/(c*x**2+b*x),x)

[Out]

2*e*sqrt(d + e*x)/c + 2*d**2*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) - 2*(b*e - c*d)**2*atan(sqrt(d + e*x)/s
qrt((b*e - c*d)/c))/(b*c**2*sqrt((b*e - c*d)/c))

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