3.9.47 \(\int \frac {15 d^2+20 d e x+8 e^2 x^2}{\sqrt {a+b x} (d+e x)^{5/2}} \, dx\) [847]

Optimal. Leaf size=116 \[ \frac {2 d^2 \sqrt {a+b x}}{(b d-a e) (d+e x)^{3/2}}+\frac {4 d (3 b d-2 a e) \sqrt {a+b x}}{(b d-a e)^2 \sqrt {d+e x}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {963, 79, 65, 223, 212} \begin {gather*} \frac {2 d^2 \sqrt {a+b x}}{(d+e x)^{3/2} (b d-a e)}+\frac {4 d \sqrt {a+b x} (3 b d-2 a e)}{\sqrt {d+e x} (b d-a e)^2}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*(d + e*x)^(5/2)),x]

[Out]

(2*d^2*Sqrt[a + b*x])/((b*d - a*e)*(d + e*x)^(3/2)) + (4*d*(3*b*d - 2*a*e)*Sqrt[a + b*x])/((b*d - a*e)^2*Sqrt[
d + e*x]) + (16*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(Sqrt[b]*Sqrt[e])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 963

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g))), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {15 d^2+20 d e x+8 e^2 x^2}{\sqrt {a+b x} (d+e x)^{5/2}} \, dx &=\frac {2 d^2 \sqrt {a+b x}}{(b d-a e) (d+e x)^{3/2}}+\frac {2 \int \frac {3 d (7 b d-6 a e)+12 e (b d-a e) x}{\sqrt {a+b x} (d+e x)^{3/2}} \, dx}{3 (b d-a e)}\\ &=\frac {2 d^2 \sqrt {a+b x}}{(b d-a e) (d+e x)^{3/2}}+\frac {4 d (3 b d-2 a e) \sqrt {a+b x}}{(b d-a e)^2 \sqrt {d+e x}}+8 \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx\\ &=\frac {2 d^2 \sqrt {a+b x}}{(b d-a e) (d+e x)^{3/2}}+\frac {4 d (3 b d-2 a e) \sqrt {a+b x}}{(b d-a e)^2 \sqrt {d+e x}}+\frac {16 \text {Subst}\left (\int \frac {1}{\sqrt {d-\frac {a e}{b}+\frac {e x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{b}\\ &=\frac {2 d^2 \sqrt {a+b x}}{(b d-a e) (d+e x)^{3/2}}+\frac {4 d (3 b d-2 a e) \sqrt {a+b x}}{(b d-a e)^2 \sqrt {d+e x}}+\frac {16 \text {Subst}\left (\int \frac {1}{1-\frac {e x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {d+e x}}\right )}{b}\\ &=\frac {2 d^2 \sqrt {a+b x}}{(b d-a e) (d+e x)^{3/2}}+\frac {4 d (3 b d-2 a e) \sqrt {a+b x}}{(b d-a e)^2 \sqrt {d+e x}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} \sqrt {e}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 99, normalized size = 0.85 \begin {gather*} \frac {2 d \sqrt {a+b x} \left (7 b d-4 a e-\frac {d e (a+b x)}{d+e x}\right )}{(b d-a e)^2 \sqrt {d+e x}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*(d + e*x)^(5/2)),x]

[Out]

(2*d*Sqrt[a + b*x]*(7*b*d - 4*a*e - (d*e*(a + b*x))/(d + e*x)))/((b*d - a*e)^2*Sqrt[d + e*x]) + (16*ArcTanh[(S
qrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(Sqrt[b]*Sqrt[e])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(600\) vs. \(2(96)=192\).
time = 0.08, size = 601, normalized size = 5.18

method	result	size
default	\(\frac {2 \sqrt {b x +a}\, \left (4 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) a^{2} e^{4} x^{2}-8 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) a b d \,e^{3} x^{2}+4 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) b^{2} d^{2} e^{2} x^{2}+8 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) a^{2} d \,e^{3} x -16 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) a b \,d^{2} e^{2} x +8 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) b^{2} d^{3} e x +4 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) a^{2} d^{2} e^{2}-8 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) a b \,d^{3} e +4 \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+a e +b d}{2 \sqrt {e b}}\right ) b^{2} d^{4}-4 a d \,e^{2} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+6 b \,d^{2} e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}-5 a \,d^{2} e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}+7 b \,d^{3} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {e b}\right )}{\sqrt {e b}\, \left (a e -b d \right )^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \left (e x +d \right )^{\frac {3}{2}}}\)	\(601\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(b*x+a)^(1/2)*(4*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*a^2*e^4*x^2-8*l
n(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*a*b*d*e^3*x^2+4*ln(1/2*(2*b*e*x+2*(
(b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*b^2*d^2*e^2*x^2+8*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))
^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*a^2*d*e^3*x-16*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+
a*e+b*d)/(e*b)^(1/2))*a*b*d^2*e^2*x+8*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/
2))*b^2*d^3*e*x+4*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*a^2*d^2*e^2-8*ln
(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*a*b*d^3*e+4*ln(1/2*(2*b*e*x+2*((b*x+
a)*(e*x+d))^(1/2)*(e*b)^(1/2)+a*e+b*d)/(e*b)^(1/2))*b^2*d^4-4*a*d*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+6*
b*d^2*e*x*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)-5*a*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(e*b)^(1/2)+7*b*d^3*((b*x+a)*(
e*x+d))^(1/2)*(e*b)^(1/2))/(e*b)^(1/2)/(a*e-b*d)^2/((b*x+a)*(e*x+d))^(1/2)/(e*x+d)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(5/2)/(b*x+a)^(1/2),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*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (100) = 200\).
time = 6.03, size = 651, normalized size = 5.61 \begin {gather*} \left [\frac {2 \, {\left (2 \, {\left (b^{2} d^{4} + a^{2} x^{2} e^{4} - 2 \, {\left (a b d x^{2} - a^{2} d x\right )} e^{3} + {\left (b^{2} d^{2} x^{2} - 4 \, a b d^{2} x + a^{2} d^{2}\right )} e^{2} + 2 \, {\left (b^{2} d^{3} x - a b d^{3}\right )} e\right )} \sqrt {b} e^{\frac {1}{2}} \log \left (b^{2} d^{2} + 4 \, {\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\frac {1}{2}} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) + {\left (7 \, b^{2} d^{3} e - 4 \, a b d x e^{3} + {\left (6 \, b^{2} d^{2} x - 5 \, a b d^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}\right )}}{b^{3} d^{4} e + a^{2} b x^{2} e^{5} - 2 \, {\left (a b^{2} d x^{2} - a^{2} b d x\right )} e^{4} + {\left (b^{3} d^{2} x^{2} - 4 \, a b^{2} d^{2} x + a^{2} b d^{2}\right )} e^{3} + 2 \, {\left (b^{3} d^{3} x - a b^{2} d^{3}\right )} e^{2}}, -\frac {2 \, {\left (4 \, {\left (b^{2} d^{4} + a^{2} x^{2} e^{4} - 2 \, {\left (a b d x^{2} - a^{2} d x\right )} e^{3} + {\left (b^{2} d^{2} x^{2} - 4 \, a b d^{2} x + a^{2} d^{2}\right )} e^{2} + 2 \, {\left (b^{2} d^{3} x - a b d^{3}\right )} e\right )} \sqrt {-b e} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {-b e} \sqrt {x e + d}}{2 \, {\left ({\left (b^{2} x^{2} + a b x\right )} e^{2} + {\left (b^{2} d x + a b d\right )} e\right )}}\right ) - {\left (7 \, b^{2} d^{3} e - 4 \, a b d x e^{3} + {\left (6 \, b^{2} d^{2} x - 5 \, a b d^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}\right )}}{b^{3} d^{4} e + a^{2} b x^{2} e^{5} - 2 \, {\left (a b^{2} d x^{2} - a^{2} b d x\right )} e^{4} + {\left (b^{3} d^{2} x^{2} - 4 \, a b^{2} d^{2} x + a^{2} b d^{2}\right )} e^{3} + 2 \, {\left (b^{3} d^{3} x - a b^{2} d^{3}\right )} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {15 d^{2} + 20 d e x + 8 e^{2} x^{2}}{\sqrt {a + b x} \left (d + e x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e**2*x**2+20*d*e*x+15*d**2)/(e*x+d)**(5/2)/(b*x+a)**(1/2),x)

[Out]

Integral((15*d**2 + 20*d*e*x + 8*e**2*x**2)/(sqrt(a + b*x)*(d + e*x)**(5/2)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (100) = 200\).
time = 3.38, size = 218, normalized size = 1.88 \begin {gather*} -\frac {16 \, \sqrt {b} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{{\left | b \right |}} + \frac {2 \, \sqrt {b x + a} {\left (\frac {2 \, {\left (3 \, b^{6} d^{2} e^{2} - 2 \, a b^{5} d e^{3}\right )} {\left (b x + a\right )}}{b^{4} d^{2} {\left | b \right |} e - 2 \, a b^{3} d {\left | b \right |} e^{2} + a^{2} b^{2} {\left | b \right |} e^{3}} + \frac {7 \, b^{7} d^{3} e - 11 \, a b^{6} d^{2} e^{2} + 4 \, a^{2} b^{5} d e^{3}}{b^{4} d^{2} {\left | b \right |} e - 2 \, a b^{3} d {\left | b \right |} e^{2} + a^{2} b^{2} {\left | b \right |} e^{3}}\right )}}{{\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-16*sqrt(b)*e^(-1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/abs(b) + 2
*sqrt(b*x + a)*(2*(3*b^6*d^2*e^2 - 2*a*b^5*d*e^3)*(b*x + a)/(b^4*d^2*abs(b)*e - 2*a*b^3*d*abs(b)*e^2 + a^2*b^2
*abs(b)*e^3) + (7*b^7*d^3*e - 11*a*b^6*d^2*e^2 + 4*a^2*b^5*d*e^3)/(b^4*d^2*abs(b)*e - 2*a*b^3*d*abs(b)*e^2 + a
^2*b^2*abs(b)*e^3))/(b^2*d + (b*x + a)*b*e - a*b*e)^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {15\,d^2+20\,d\,e\,x+8\,e^2\,x^2}{\sqrt {a+b\,x}\,{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*d^2 + 8*e^2*x^2 + 20*d*e*x)/((a + b*x)^(1/2)*(d + e*x)^(5/2)),x)

[Out]

int((15*d^2 + 8*e^2*x^2 + 20*d*e*x)/((a + b*x)^(1/2)*(d + e*x)^(5/2)), x)

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