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

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

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Rubi [A]  time = 0.12, 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 = {949, 78, 63, 217, 206} \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 63

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 78

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] || LtQ
[p, n]))))

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 949

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 \operatorname {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 \operatorname {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 = 0.30, size = 128, normalized size = 1.10 \begin {gather*} \frac {2 \left (\frac {8 (b d-a e)^{3/2} \left (\frac {b (d+e x)}{b d-a e}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{b^2 \sqrt {e}}+\frac {d \sqrt {a+b x} (b d (7 d+6 e x)-a e (5 d+4 e x))}{(b d-a e)^2}\right )}{(d+e x)^{3/2}} \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]*(-(a*e*(5*d + 4*e*x)) + b*d*(7*d + 6*e*x)))/(b*d - a*e)^2 + (8*(b*d - a*e)^(3/2)*((b*(d +
e*x))/(b*d - a*e))^(3/2)*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/(b^2*Sqrt[e])))/(d + e*x)^(3/2)

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

IntegrateAlgebraic[(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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fricas [B]  time = 0.60, size = 665, normalized size = 5.73 \begin {gather*} \left [\frac {2 \, {\left (2 \, {\left (b^{2} d^{4} - 2 \, a b d^{3} e + a^{2} d^{2} e^{2} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3}\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + {\left (7 \, b^{2} d^{3} e - 5 \, a b d^{2} e^{2} + 2 \, {\left (3 \, b^{2} d^{2} e^{2} - 2 \, a b d e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}\right )}}{b^{3} d^{4} e - 2 \, a b^{2} d^{3} e^{2} + a^{2} b d^{2} e^{3} + {\left (b^{3} d^{2} e^{3} - 2 \, a b^{2} d e^{4} + a^{2} b e^{5}\right )} x^{2} + 2 \, {\left (b^{3} d^{3} e^{2} - 2 \, a b^{2} d^{2} e^{3} + a^{2} b d e^{4}\right )} x}, -\frac {2 \, {\left (4 \, {\left (b^{2} d^{4} - 2 \, a b d^{3} e + a^{2} d^{2} e^{2} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3}\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - {\left (7 \, b^{2} d^{3} e - 5 \, a b d^{2} e^{2} + 2 \, {\left (3 \, b^{2} d^{2} e^{2} - 2 \, a b d e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}\right )}}{b^{3} d^{4} e - 2 \, a b^{2} d^{3} e^{2} + a^{2} b d^{2} e^{3} + {\left (b^{3} d^{2} e^{3} - 2 \, a b^{2} d e^{4} + a^{2} b e^{5}\right )} x^{2} + 2 \, {\left (b^{3} d^{3} e^{2} - 2 \, a b^{2} d^{2} e^{3} + a^{2} b d e^{4}\right )} x}\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 - 2*a*b*d^3*e + a^2*d^2*e^2 + (b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*x^2 + 2*(b^2*d^3*e - 2*a*b*
d^2*e^2 + a^2*d*e^3)*x)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*
sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) + (7*b^2*d^3*e - 5*a*b*d^2*e^2 + 2*(3*b^2*d^2
*e^2 - 2*a*b*d*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*d^4*e - 2*a*b^2*d^3*e^2 + a^2*b*d^2*e^3 + (b^3*d^2*e^
3 - 2*a*b^2*d*e^4 + a^2*b*e^5)*x^2 + 2*(b^3*d^3*e^2 - 2*a*b^2*d^2*e^3 + a^2*b*d*e^4)*x), -2*(4*(b^2*d^4 - 2*a*
b*d^3*e + a^2*d^2*e^2 + (b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*x^2 + 2*(b^2*d^3*e - 2*a*b*d^2*e^2 + a^2*d*e^3)*
x)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e +
(b^2*d*e + a*b*e^2)*x)) - (7*b^2*d^3*e - 5*a*b*d^2*e^2 + 2*(3*b^2*d^2*e^2 - 2*a*b*d*e^3)*x)*sqrt(b*x + a)*sqr
t(e*x + d))/(b^3*d^4*e - 2*a*b^2*d^3*e^2 + a^2*b*d^2*e^3 + (b^3*d^2*e^3 - 2*a*b^2*d*e^4 + a^2*b*e^5)*x^2 + 2*(
b^3*d^3*e^2 - 2*a*b^2*d^2*e^3 + a^2*b*d*e^4)*x)]

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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((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]

Timed out

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maple [B]  time = 0.03, size = 601, normalized size = 5.18 \begin {gather*} \frac {2 \sqrt {b x +a}\, \left (4 a^{2} e^{4} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-8 a b d \,e^{3} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+4 b^{2} d^{2} e^{2} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+8 a^{2} d \,e^{3} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-16 a b \,d^{2} e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+8 b^{2} d^{3} e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+4 a^{2} d^{2} e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-8 a b \,d^{3} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+4 b^{2} d^{4} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, a d \,e^{2} x +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b \,d^{2} e x -5 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, a \,d^{2} e +7 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b \,d^{3}\right )}{\sqrt {b e}\, \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}}} \end {gather*}

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)

[Out]

2*(b*x+a)^(1/2)*(4*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*a^2*e^4-8*l
n(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*a*b*d*e^3+4*ln(1/2*(2*b*e*x+a*e
+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*b^2*d^2*e^2+8*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*a^2*d*e^3-16*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e
)^(1/2))/(b*e)^(1/2))*x*a*b*d^2*e^2+8*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/
2))*x*b^2*d^3*e+4*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a^2*d^2*e^2-8*ln
(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a*b*d^3*e+4*ln(1/2*(2*b*e*x+a*e+b*d+
2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*b^2*d^4-4*x*a*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*
x*b*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-5*a*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+7*b*d^3*((b*x+a)*(
e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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

[Out]

Timed out

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