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

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

[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])

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Rubi [A]  time = 0.121351, antiderivative size = 116, normalized size of antiderivative = 1., 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} \[ \frac{2 d^2 \sqrt{a+b x}}{(d+e x)^{3/2} (b d-a e)}+\frac{4 d \sqrt{a+b x} (3 b d-2 a e)}{\sqrt{d+e x} (b d-a e)^2}+\frac{16 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{\sqrt{b} \sqrt{e}} \]

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 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]

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 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 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 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])

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*}

Mathematica [A]  time = 0.304688, size = 128, normalized size = 1.1 \[ \frac{2 \left (\frac{8 (b d-a e)^{3/2} \left (\frac{b (d+e x)}{b d-a e}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{b^2 \sqrt{e}}+\frac{d \sqrt{a+b x} (b d (7 d+6 e x)-a e (5 d+4 e x))}{(b d-a e)^2}\right )}{(d+e x)^{3/2}} \]

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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Maple [B]  time = 0.345, size = 601, normalized size = 5.2 \begin{align*} 2\,{\frac{\sqrt{bx+a}}{\sqrt{be} \left ( ae-bd \right ) ^{2}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) } \left ( ex+d \right ) ^{3/2}} \left ( 4\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){x}^{2}{a}^{2}{e}^{4}-8\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){x}^{2}abd{e}^{3}+4\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){x}^{2}{b}^{2}{d}^{2}{e}^{2}+8\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) x{a}^{2}d{e}^{3}-16\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) xab{d}^{2}{e}^{2}+8\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) x{b}^{2}{d}^{3}e+4\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}{d}^{2}{e}^{2}-8\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) ab{d}^{3}e+4\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{2}{d}^{4}-4\,xad{e}^{2}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+6\,xb{d}^{2}e\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}-5\,a{d}^{2}e\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+7\,b{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be} \right ) } \end{align*}

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*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*a^2*e^4-8*l
n(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*a*b*d*e^3+4*ln(1/2*(2*b*x*e+2*(
(b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*b^2*d^2*e^2+8*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))
^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a^2*d*e^3-16*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+
a*e+b*d)/(b*e)^(1/2))*x*a*b*d^2*e^2+8*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/
2))*x*b^2*d^3*e+4*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*d^2*e^2-8*ln
(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d^3*e+4*ln(1/2*(2*b*x*e+2*((b*x+
a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 3.35777, size = 1407, normalized size = 12.13 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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]  time = 1.29966, size = 294, normalized size = 2.53 \begin{align*} -\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{align*}

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)