3.5.50 \(\int \frac {\sqrt {a+c x^2}}{(d+e x)^2} \, dx\)

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

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Rubi [A]  time = 0.07, 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 = {733, 844, 217, 206, 725} \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 206

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

Rule 217

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 725

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 733

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 844

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 \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{e^2}+\frac {(c d) \operatorname {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 = 0.13, size = 134, normalized size = 1.22 \begin {gather*} \frac {\frac {c d \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )}{\sqrt {a e^2+c d^2}}-\frac {c d \log (d+e x)}{\sqrt {a e^2+c d^2}}-\frac {e \sqrt {a+c x^2}}{d+e x}+\sqrt {c} \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\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)) - (c*d*Log[d + e*x])/Sqrt[c*d^2 + a*e^2] + Sqrt[c]*Log[c*x + Sqrt[c]*Sqrt[a
+ c*x^2]] + (c*d*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/Sqrt[c*d^2 + a*e^2])/e^2

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IntegrateAlgebraic [A]  time = 0.64, size = 174, normalized size = 1.58 \begin {gather*} -\frac {2 c d \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{e^2 \left (a e^2+c d^2\right )}-\frac {\sqrt {a+c x^2}}{e (d+e x)}-\frac {\sqrt {c} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 884, normalized size = 8.04 \begin {gather*} \left [\frac {{\left (c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + {\left (c d e x + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c d^{3} e^{2} + a d e^{4} + {\left (c d^{2} e^{3} + a e^{5}\right )} x\right )}}, \frac {2 \, {\left (c d e x + 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}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\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^{3} e^{2} + a d e^{4} + {\left (c d^{2} e^{3} + a e^{5}\right )} x\right )}}, -\frac {2 \, {\left (c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (c d e x + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c d^{3} e^{2} + a d e^{4} + {\left (c d^{2} e^{3} + a e^{5}\right )} x\right )}}, \frac {{\left (c d e x + 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}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\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^{3} e^{2} + a d e^{4} + {\left (c d^{2} e^{3} + a e^{5}\right )} x}\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^3 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (c*d*e
*x + c*d^2)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*
d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*(c*d^2*e + a*e^3)*sqrt(c*x^2 + a))/
(c*d^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*x), 1/2*(2*(c*d*e*x + 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)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) + (c*d^3 + a*d*e^2 + (
c*d^2*e + a*e^3)*x)*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^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*x), -1/2*(2*(c*d^3 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(-c)*a
rctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (c*d*e*x + c*d^2)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e
^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2
)) + 2*(c*d^2*e + a*e^3)*sqrt(c*x^2 + a))/(c*d^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*x), ((c*d*e*x + c*d^2)*sq
rt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a
*c*e^2)*x^2)) - (c*d^3 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*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^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*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,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Error: Bad Argument Type

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maple [B]  time = 0.06, size = 649, normalized size = 5.90 \begin {gather*} \frac {a c d \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e}+\frac {c^{2} d^{3} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{3}}+\frac {a \sqrt {c}\, \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{a \,e^{2}+c \,d^{2}}+\frac {c^{\frac {3}{2}} d^{2} \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) e^{2}}+\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, c x}{a \,e^{2}+c \,d^{2}}-\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, c d}{\left (a \,e^{2}+c \,d^{2}\right ) e}-\frac {\left (-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.56, size = 94, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {c x^{2} + a}}{e^{2} x + d e} + \frac {\sqrt {c} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{2}} - \frac {c d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{\sqrt {a + \frac {c d^{2}}{e^{2}}} e^{3}} \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]

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

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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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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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