3.6.30 \(\int \frac {\sqrt {a+c x^2}}{(d+e x)^2} \, dx\) [530]

Optimal. Leaf size=110 \[ -\frac {\sqrt {a+c x^2}}{e (d+e x)}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{e^2}+\frac {c d \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^2 \sqrt {c d^2+a e^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {747, 858, 223, 212, 739} \begin {gather*} \frac {c d \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^2 \sqrt {a e^2+c d^2}}-\frac {\sqrt {a+c x^2}}{e (d+e x)}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^2]/(d + e*x)^2,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 747

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 1))), x] - Dist[2*c*(p/(e*(m + 1))), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.48, size = 119, normalized size = 1.08 \begin {gather*} -\frac {\frac {e \sqrt {a+c x^2}}{d+e x}-\frac {2 c d \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\sqrt {-c d^2-a e^2}}+\sqrt {c} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^2]/(d + e*x)^2,x]

[Out]

-(((e*Sqrt[a + c*x^2])/(d + e*x) - (2*c*d*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2
]])/Sqrt[-(c*d^2) - a*e^2] + Sqrt[c]*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/e^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(535\) vs. \(2(96)=192\).
time = 0.44, size = 536, normalized size = 4.87

method	result	size
default	\(\frac {-\frac {e^{2} \left (c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {c d e \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{2} a +c \,d^{2}}+\frac {2 c \,e^{2} \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{e^{2} a +c \,d^{2}}}{e^{2}}\)	\(536\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(1/2)/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/e^2*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(3/2)-c*d*e/(a*e^2+c*d^2)*
((c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-c^(1/2)*d/e*ln((-c*d/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2-
2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))-(a*e^2+c*d^2)/e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-
2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))+2
*c/(a*e^2+c*d^2)*e^2*(1/4*(2*c*(x+d/e)-2*c*d/e)/c*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)+1/8*(4
*c*(a*e^2+c*d^2)/e^2-4*c^2*d^2/e^2)/c^(3/2)*ln((-c*d/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+
c*d^2)/e^2)^(1/2))))

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Maxima [A]
time = 0.32, size = 94, normalized size = 0.85 \begin {gather*} -\frac {c d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | x e + d \right |}} - \frac {a e}{\sqrt {a c} {\left | x e + d \right |}}\right ) e^{\left (-3\right )}}{\sqrt {c d^{2} e^{\left (-2\right )} + a}} + \sqrt {c} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-2\right )} - \frac {\sqrt {c x^{2} + a}}{x e^{2} + d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^2,x, algorithm="maxima")

[Out]

-c*d*arcsinh(c*d*x/(sqrt(a*c)*abs(x*e + d)) - a*e/(sqrt(a*c)*abs(x*e + d)))*e^(-3)/sqrt(c*d^2*e^(-2) + a) + sq
rt(c)*arcsinh(c*x/sqrt(a*c))*e^(-2) - sqrt(c*x^2 + a)/(x*e^2 + d*e)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (94) = 188\).
time = 2.37, size = 852, normalized size = 7.75 \begin {gather*} \left [\frac {{\left (c d^{2} x e + c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + {\left (c d x e + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c d^{2} x e^{3} + c d^{3} e^{2} + a x e^{5} + a d e^{4}\right )}}, -\frac {2 \, {\left (c d x e + c d^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) - {\left (c d^{2} x e + c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c d^{2} x e^{3} + c d^{3} e^{2} + a x e^{5} + a d e^{4}\right )}}, -\frac {2 \, {\left (c d^{2} x e + c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (c d x e + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c d^{2} x e^{3} + c d^{3} e^{2} + a x e^{5} + a d e^{4}\right )}}, -\frac {{\left (c d x e + c d^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + {\left (c d^{2} x e + c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{c d^{2} x e^{3} + c d^{3} e^{2} + a x e^{5} + a d e^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

[1/2*((c*d^2*x*e + c*d^3 + a*x*e^3 + a*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (c*d*x
*e + c*d^2)*sqrt(c*d^2 + a*e^2)*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a
*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) - 2*(c*d^2*e + a*e^3)*sqrt(c*x^2 + a))
/(c*d^2*x*e^3 + c*d^3*e^2 + a*x*e^5 + a*d*e^4), -1/2*(2*(c*d*x*e + c*d^2)*sqrt(-c*d^2 - a*e^2)*arctan(-sqrt(-c
*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e^2)) - (c*d^2*x*e + c*d^
3 + a*x*e^3 + a*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(c*d^2*e + a*e^3)*sqrt(c*x^
2 + a))/(c*d^2*x*e^3 + c*d^3*e^2 + a*x*e^5 + a*d*e^4), -1/2*(2*(c*d^2*x*e + c*d^3 + a*x*e^3 + a*d*e^2)*sqrt(-c
)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (c*d*x*e + c*d^2)*sqrt(c*d^2 + a*e^2)*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e
+ a*c*d^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e +
d^2)) + 2*(c*d^2*e + a*e^3)*sqrt(c*x^2 + a))/(c*d^2*x*e^3 + c*d^3*e^2 + a*x*e^5 + a*d*e^4), -((c*d*x*e + c*d^
2)*sqrt(-c*d^2 - a*e^2)*arctan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a
*c*x^2 + a^2)*e^2)) + (c*d^2*x*e + c*d^3 + a*x*e^3 + a*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (c
*d^2*e + a*e^3)*sqrt(c*x^2 + a))/(c*d^2*x*e^3 + c*d^3*e^2 + a*x*e^5 + a*d*e^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + c x^{2}}}{\left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(1/2)/(e*x+d)**2,x)

[Out]

Integral(sqrt(a + c*x**2)/(d + e*x)**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+a}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(1/2)/(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(1/2)/(d + e*x)^2, x)

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