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

Optimal. Leaf size=153 \[ \frac {3 \sqrt {c} \left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 e^4}+\frac {3 c d \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4}-\frac {3 c \sqrt {a+c x^2} (2 d-e x)}{2 e^3}-\frac {\left (a+c x^2\right )^{3/2}}{e (d+e x)} \]

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Rubi [A]  time = 0.14, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {733, 815, 844, 217, 206, 725} \begin {gather*} \frac {3 \sqrt {c} \left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 e^4}+\frac {3 c d \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4}-\frac {3 c \sqrt {a+c x^2} (2 d-e x)}{2 e^3}-\frac {\left (a+c x^2\right )^{3/2}}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^(3/2)/(d + e*x)^2,x]

[Out]

(-3*c*(2*d - e*x)*Sqrt[a + c*x^2])/(2*e^3) - (a + c*x^2)^(3/2)/(e*(d + e*x)) + (3*Sqrt[c]*(2*c*d^2 + a*e^2)*Ar
cTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*e^4) + (3*c*d*Sqrt[c*d^2 + a*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a
*e^2]*Sqrt[a + c*x^2])])/e^4

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 815

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

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

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Mathematica [A]  time = 0.17, size = 179, normalized size = 1.17 \begin {gather*} \frac {-\frac {e \sqrt {a+c x^2} \left (2 a e^2+c \left (6 d^2+3 d e x-e^2 x^2\right )\right )}{d+e x}+3 \sqrt {c} \left (a e^2+2 c d^2\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )+6 c d \sqrt {a e^2+c d^2} \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )-6 c d \sqrt {a e^2+c d^2} \log (d+e x)}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((e*Sqrt[a + c*x^2]*(2*a*e^2 + c*(6*d^2 + 3*d*e*x - e^2*x^2)))/(d + e*x)) - 6*c*d*Sqrt[c*d^2 + a*e^2]*Log[d
+ e*x] + 3*Sqrt[c]*(2*c*d^2 + a*e^2)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]] + 6*c*d*Sqrt[c*d^2 + a*e^2]*Log[a*e -
c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(2*e^4)

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IntegrateAlgebraic [A]  time = 0.96, size = 208, normalized size = 1.36 \begin {gather*} -\frac {3 \left (a \sqrt {c} e^2+2 c^{3/2} d^2\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 e^4}-\frac {6 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^4}+\frac {\sqrt {a+c x^2} \left (-2 a e^2-6 c d^2-3 c d e x+c e^2 x^2\right )}{2 e^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + c*x^2)^(3/2)/(d + e*x)^2,x]

[Out]

(Sqrt[a + c*x^2]*(-6*c*d^2 - 2*a*e^2 - 3*c*d*e*x + c*e^2*x^2))/(2*e^3*(d + e*x)) - (6*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^4 - (3*(2*c^(3/2)*d^2 + a*Sqrt[c]*e^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(2*e^4)

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fricas [A]  time = 0.72, size = 888, normalized size = 5.80 \begin {gather*} \left [\frac {3 \, {\left (2 \, c d^{3} + a d e^{2} + {\left (2 \, 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 ) + 6 \, {\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 e^{3} x^{2} - 3 \, c d e^{2} x - 6 \, c d^{2} e - 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (e^{5} x + d e^{4}\right )}}, \frac {12 \, {\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 ) + 3 \, {\left (2 \, c d^{3} + a d e^{2} + {\left (2 \, 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 e^{3} x^{2} - 3 \, c d e^{2} x - 6 \, c d^{2} e - 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (e^{5} x + d e^{4}\right )}}, -\frac {3 \, {\left (2 \, c d^{3} + a d e^{2} + {\left (2 \, c d^{2} e + a e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 3 \, {\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 ) - {\left (c e^{3} x^{2} - 3 \, c d e^{2} x - 6 \, c d^{2} e - 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (e^{5} x + d e^{4}\right )}}, \frac {6 \, {\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 ) - 3 \, {\left (2 \, c d^{3} + a d e^{2} + {\left (2 \, 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 e^{3} x^{2} - 3 \, c d e^{2} x - 6 \, c d^{2} e - 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (e^{5} x + d e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(2*c*d^3 + a*d*e^2 + (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) +
6*(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*e^3*x^2 - 3*c*d*e^2*x -
6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4), 1/4*(12*(c*d*e*x + c*d^2)*sqrt(-c*d^2 - a*e^2)*arctan(s
qrt(-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)) + 3*(2*c*d^3
+ a*d*e^2 + (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*e^3*x^2 - 3*
c*d*e^2*x - 6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4), -1/2*(3*(2*c*d^3 + a*d*e^2 + (2*c*d^2*e + a
*e^3)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 3*(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)) - (c*e^3*x^2 - 3*c*d*e^2*x - 6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4), 1/
2*(6*(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)) - 3*(2*c*d^3 + a*d*e^2 + (2*c*d^2*e + a*e^3)*x)*sqrt(-c)*arctan(sqrt(-c
)*x/sqrt(c*x^2 + a)) + (c*e^3*x^2 - 3*c*d*e^2*x - 6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 1154, normalized size = 7.54

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(3/2)/(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)^(5/2)-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)^(3/2)+3/2/e^2*c^2*d^2/(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)*x+9/2/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))*a-3/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)*a-3/e^3*c^2*d^3/(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)+3/e^4*c^(5/2)*d^4/(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))+3/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^2+6/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))*a+3/e^5*c^3*d^5/(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)^(3/2)*x+3/2/(a*e^2+c*d^2)*c*a*(-2*(x+d/e)*c
*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)*x+3/2/(a*e^2+c*d^2)*c^(1/2)*a^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))

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maxima [A]  time = 1.68, size = 148, normalized size = 0.97 \begin {gather*} -\frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}}}{e^{2} x + d e} + \frac {3 \, \sqrt {c x^{2} + a} c x}{2 \, e^{2}} + \frac {3 \, c^{\frac {3}{2}} d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{4}} + \frac {3 \, a \sqrt {c} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, e^{2}} - \frac {3 \, \sqrt {a + \frac {c d^{2}}{e^{2}}} 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 )}{e^{3}} - \frac {3 \, \sqrt {c x^{2} + a} c d}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*x^2 + a)^(3/2)/(e^2*x + d*e) + 3/2*sqrt(c*x^2 + a)*c*x/e^2 + 3*c^(3/2)*d^2*arcsinh(c*x/sqrt(a*c))/e^4 + 3/
2*a*sqrt(c)*arcsinh(c*x/sqrt(a*c))/e^2 - 3*sqrt(a + c*d^2/e^2)*c*d*arcsinh(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*
e/(sqrt(a*c)*abs(e*x + d)))/e^3 - 3*sqrt(c*x^2 + a)*c*d/e^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}}{{\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)^(3/2)/(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(3/2)/(d + e*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + c x^{2}\right )^{\frac {3}{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)**(3/2)/(e*x+d)**2,x)

[Out]

Integral((a + c*x**2)**(3/2)/(d + e*x)**2, x)

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