3.5.87 \(\int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac {3 c d e \sqrt {a+c x^2}}{2 (d+e x) \left (a e^2+c d^2\right )^2}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac {c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {745, 807, 725, 206} \begin {gather*} -\frac {3 c d e \sqrt {a+c x^2}}{2 (d+e x) \left (a e^2+c d^2\right )^2}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac {c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*Sqrt[a + c*x^2]),x]

[Out]

-(e*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)*(d + e*x)^2) - (3*c*d*e*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)^2*(d + e*x
)) - (c*(2*c*d^2 - a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(2*(c*d^2 + a*e^2)^(5/
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 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 745

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 161, normalized size = 1.11 \begin {gather*} \frac {-e \sqrt {a+c x^2} \sqrt {a e^2+c d^2} \left (a e^2+c d (4 d+3 e x)\right )-c (d+e x)^2 \left (2 c d^2-a e^2\right ) \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )+c (d+e x)^2 \left (2 c d^2-a e^2\right ) \log (d+e x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*Sqrt[a + c*x^2]),x]

[Out]

(-(e*Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]*(a*e^2 + c*d*(4*d + 3*e*x))) + c*(2*c*d^2 - a*e^2)*(d + e*x)^2*Log[d
+ e*x] - c*(2*c*d^2 - a*e^2)*(d + e*x)^2*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(2*(c*d^2 + a
*e^2)^(5/2)*(d + e*x)^2)

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IntegrateAlgebraic [A]  time = 0.93, size = 171, normalized size = 1.18 \begin {gather*} \frac {\left (2 c^2 d^2 \sqrt {-a e^2-c d^2}-a c e^2 \sqrt {-a e^2-c d^2}\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}\right )}{\left (a e^2+c d^2\right )^3}+\frac {\sqrt {a+c x^2} \left (-a e^3-4 c d^2 e-3 c d e^2 x\right )}{2 (d+e x)^2 \left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^3*Sqrt[a + c*x^2]),x]

[Out]

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

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fricas [B]  time = 0.67, size = 713, normalized size = 4.92 \begin {gather*} \left [-\frac {{\left (2 \, c^{2} d^{4} - a c d^{2} e^{2} + {\left (2 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 2 \, {\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x\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 (4 \, c^{2} d^{4} e + 5 \, a c d^{2} e^{3} + a^{2} e^{5} + 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{4 \, {\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}, -\frac {{\left (2 \, c^{2} d^{4} - a c d^{2} e^{2} + {\left (2 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 2 \, {\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x\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 (4 \, c^{2} d^{4} e + 5 \, a c d^{2} e^{3} + a^{2} e^{5} + 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((2*c^2*d^4 - a*c*d^2*e^2 + (2*c^2*d^2*e^2 - a*c*e^4)*x^2 + 2*(2*c^2*d^3*e - a*c*d*e^3)*x)*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*(4*c^2*d^4*e + 5*a*c*d^2*e^3 + a^2*e^5 + 3*(c^2*d^3*e^2 + a*
c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^8 + 3*a*c^2*d^6*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c
^2*d^4*e^4 + 3*a^2*c*d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x)
, -1/2*((2*c^2*d^4 - a*c*d^2*e^2 + (2*c^2*d^2*e^2 - a*c*e^4)*x^2 + 2*(2*c^2*d^3*e - a*c*d*e^3)*x)*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)) + (4*c^2*d^4*e + 5*a*c*d^2*e^3 + a^2*e^5 + 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^8 + 3*a*
c^2*d^6*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 + a^3*e^8)*x^2
+ 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x)]

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giac [B]  time = 0.28, size = 345, normalized size = 2.38 \begin {gather*} -c {\left (\frac {{\left (2 \, c d^{2} - a e^{2}\right )} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} c d^{2} e + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} c^{\frac {3}{2}} d^{3} - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a c d^{2} e - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a \sqrt {c} d e^{2} - {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} a e^{3} + 3 \, a^{2} \sqrt {c} d e^{2} - {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a^{2} e^{3}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} e + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} \sqrt {c} d - a e\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-c*((2*c*d^2 - a*e^2)*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((c^2*d^4 + 2
*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c*d^2 - a*e^2)) + (2*(sqrt(c)*x - sqrt(c*x^2 + a))^3*c*d^2*e + 6*(sqrt(c)*x - sq
rt(c*x^2 + a))^2*c^(3/2)*d^3 - 10*(sqrt(c)*x - sqrt(c*x^2 + a))*a*c*d^2*e - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^2*
a*sqrt(c)*d*e^2 - (sqrt(c)*x - sqrt(c*x^2 + a))^3*a*e^3 + 3*a^2*sqrt(c)*d*e^2 - (sqrt(c)*x - sqrt(c*x^2 + a))*
a^2*e^3)/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*((sqrt(c)*x - sqrt(c*x^2 + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 +
a))*sqrt(c)*d - a*e)^2))

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maple [B]  time = 0.06, size = 426, normalized size = 2.94 \begin {gather*} -\frac {3 c^{2} d^{2} \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 )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e}-\frac {3 \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}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (x +\frac {d}{e}\right )}+\frac {c \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 )}{2 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e}-\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}}}}{2 \left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/(c*x^2+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.64, size = 243, normalized size = 1.68 \begin {gather*} -\frac {3 \, \sqrt {c x^{2} + a} c d}{2 \, {\left (c^{2} d^{4} x + 2 \, a c d^{2} e^{2} x + a^{2} e^{4} x + \frac {c^{2} d^{5}}{e} + 2 \, a c d^{3} e + a^{2} d e^{3}\right )}} - \frac {\sqrt {c x^{2} + a}}{2 \, {\left (c d^{2} e x^{2} + a e^{3} x^{2} + 2 \, c d^{3} x + 2 \, a d e^{2} x + \frac {c d^{4}}{e} + a d^{2} e\right )}} + \frac {3 \, c^{2} d^{2} \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 )}{2 \, {\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {5}{2}} e^{5}} - \frac {c \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 )}{2 \, {\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

-3/2*sqrt(c*x^2 + a)*c*d/(c^2*d^4*x + 2*a*c*d^2*e^2*x + a^2*e^4*x + c^2*d^5/e + 2*a*c*d^3*e + a^2*d*e^3) - 1/2
*sqrt(c*x^2 + a)/(c*d^2*e*x^2 + a*e^3*x^2 + 2*c*d^3*x + 2*a*d*e^2*x + c*d^4/e + a*d^2*e) + 3/2*c^2*d^2*arcsinh
(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/((a + c*d^2/e^2)^(5/2)*e^5) - 1/2*c*arcsinh(c*
d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/((a + c*d^2/e^2)^(3/2)*e^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)^(1/2)*(d + e*x)^3),x)

[Out]

int(1/((a + c*x^2)^(1/2)*(d + e*x)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + c x^{2}} \left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(sqrt(a + c*x**2)*(d + e*x)**3), x)

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