3.849 \(\int \frac {15 d^2+20 d e x+8 e^2 x^2}{\sqrt {a+b x} (d+e x)^{9/2}} \, dx\)

Optimal. Leaf size=189 \[ \frac {32 b \sqrt {a+b x} \left (35 a^2 e^2-84 a b d e+58 b^2 d^2\right )}{105 \sqrt {d+e x} (b d-a e)^4}+\frac {16 \sqrt {a+b x} \left (35 a^2 e^2-84 a b d e+58 b^2 d^2\right )}{105 (d+e x)^{3/2} (b d-a e)^3}+\frac {6 d^2 \sqrt {a+b x}}{7 (d+e x)^{7/2} (b d-a e)}+\frac {4 d \sqrt {a+b x} (23 b d-14 a e)}{35 (d+e x)^{5/2} (b d-a e)^2} \]

[Out]

6/7*d^2*(b*x+a)^(1/2)/(-a*e+b*d)/(e*x+d)^(7/2)+4/35*d*(-14*a*e+23*b*d)*(b*x+a)^(1/2)/(-a*e+b*d)^2/(e*x+d)^(5/2
)+16/105*(35*a^2*e^2-84*a*b*d*e+58*b^2*d^2)*(b*x+a)^(1/2)/(-a*e+b*d)^3/(e*x+d)^(3/2)+32/105*b*(35*a^2*e^2-84*a
*b*d*e+58*b^2*d^2)*(b*x+a)^(1/2)/(-a*e+b*d)^4/(e*x+d)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {949, 78, 45, 37} \[ \frac {32 b \sqrt {a+b x} \left (35 a^2 e^2-84 a b d e+58 b^2 d^2\right )}{105 \sqrt {d+e x} (b d-a e)^4}+\frac {16 \sqrt {a+b x} \left (35 a^2 e^2-84 a b d e+58 b^2 d^2\right )}{105 (d+e x)^{3/2} (b d-a e)^3}+\frac {6 d^2 \sqrt {a+b x}}{7 (d+e x)^{7/2} (b d-a e)}+\frac {4 d \sqrt {a+b x} (23 b d-14 a e)}{35 (d+e x)^{5/2} (b d-a e)^2} \]

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)^(9/2)),x]

[Out]

(6*d^2*Sqrt[a + b*x])/(7*(b*d - a*e)*(d + e*x)^(7/2)) + (4*d*(23*b*d - 14*a*e)*Sqrt[a + b*x])/(35*(b*d - a*e)^
2*(d + e*x)^(5/2)) + (16*(58*b^2*d^2 - 84*a*b*d*e + 35*a^2*e^2)*Sqrt[a + b*x])/(105*(b*d - a*e)^3*(d + e*x)^(3
/2)) + (32*b*(58*b^2*d^2 - 84*a*b*d*e + 35*a^2*e^2)*Sqrt[a + b*x])/(105*(b*d - a*e)^4*Sqrt[d + e*x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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 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)^{9/2}} \, dx &=\frac {6 d^2 \sqrt {a+b x}}{7 (b d-a e) (d+e x)^{7/2}}+\frac {2 \int \frac {3 d (17 b d-14 a e)+28 e (b d-a e) x}{\sqrt {a+b x} (d+e x)^{7/2}} \, dx}{7 (b d-a e)}\\ &=\frac {6 d^2 \sqrt {a+b x}}{7 (b d-a e) (d+e x)^{7/2}}+\frac {4 d (23 b d-14 a e) \sqrt {a+b x}}{35 (b d-a e)^2 (d+e x)^{5/2}}+\frac {\left (8 \left (58 b^2 d^2-84 a b d e+35 a^2 e^2\right )\right ) \int \frac {1}{\sqrt {a+b x} (d+e x)^{5/2}} \, dx}{35 (b d-a e)^2}\\ &=\frac {6 d^2 \sqrt {a+b x}}{7 (b d-a e) (d+e x)^{7/2}}+\frac {4 d (23 b d-14 a e) \sqrt {a+b x}}{35 (b d-a e)^2 (d+e x)^{5/2}}+\frac {16 \left (58 b^2 d^2-84 a b d e+35 a^2 e^2\right ) \sqrt {a+b x}}{105 (b d-a e)^3 (d+e x)^{3/2}}+\frac {\left (16 b \left (58 b^2 d^2-84 a b d e+35 a^2 e^2\right )\right ) \int \frac {1}{\sqrt {a+b x} (d+e x)^{3/2}} \, dx}{105 (b d-a e)^3}\\ &=\frac {6 d^2 \sqrt {a+b x}}{7 (b d-a e) (d+e x)^{7/2}}+\frac {4 d (23 b d-14 a e) \sqrt {a+b x}}{35 (b d-a e)^2 (d+e x)^{5/2}}+\frac {16 \left (58 b^2 d^2-84 a b d e+35 a^2 e^2\right ) \sqrt {a+b x}}{105 (b d-a e)^3 (d+e x)^{3/2}}+\frac {32 b \left (58 b^2 d^2-84 a b d e+35 a^2 e^2\right ) \sqrt {a+b x}}{105 (b d-a e)^4 \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 173, normalized size = 0.92 \[ \frac {2 \sqrt {a+b x} \left (-a^3 e^3 \left (409 d^2+644 d e x+280 e^2 x^2\right )+a^2 b e^2 \left (1953 d^3+3890 d^2 e x+2632 d e^2 x^2+560 e^3 x^3\right )-a b^2 d e \left (2975 d^3+6664 d^2 e x+5168 d e^2 x^2+1344 e^3 x^3\right )+b^3 d^2 \left (1575 d^3+3850 d^2 e x+3248 d e^2 x^2+928 e^3 x^3\right )\right )}{105 (d+e x)^{7/2} (b d-a e)^4} \]

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)^(9/2)),x]

[Out]

(2*Sqrt[a + b*x]*(-(a^3*e^3*(409*d^2 + 644*d*e*x + 280*e^2*x^2)) + a^2*b*e^2*(1953*d^3 + 3890*d^2*e*x + 2632*d
*e^2*x^2 + 560*e^3*x^3) + b^3*d^2*(1575*d^3 + 3850*d^2*e*x + 3248*d*e^2*x^2 + 928*e^3*x^3) - a*b^2*d*e*(2975*d
^3 + 6664*d^2*e*x + 5168*d*e^2*x^2 + 1344*e^3*x^3)))/(105*(b*d - a*e)^4*(d + e*x)^(7/2))

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fricas [B]  time = 6.62, size = 487, normalized size = 2.58 \[ \frac {2 \, {\left (1575 \, b^{3} d^{5} - 2975 \, a b^{2} d^{4} e + 1953 \, a^{2} b d^{3} e^{2} - 409 \, a^{3} d^{2} e^{3} + 16 \, {\left (58 \, b^{3} d^{2} e^{3} - 84 \, a b^{2} d e^{4} + 35 \, a^{2} b e^{5}\right )} x^{3} + 8 \, {\left (406 \, b^{3} d^{3} e^{2} - 646 \, a b^{2} d^{2} e^{3} + 329 \, a^{2} b d e^{4} - 35 \, a^{3} e^{5}\right )} x^{2} + 2 \, {\left (1925 \, b^{3} d^{4} e - 3332 \, a b^{2} d^{3} e^{2} + 1945 \, a^{2} b d^{2} e^{3} - 322 \, a^{3} d e^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{105 \, {\left (b^{4} d^{8} - 4 \, a b^{3} d^{7} e + 6 \, a^{2} b^{2} d^{6} e^{2} - 4 \, a^{3} b d^{5} e^{3} + a^{4} d^{4} e^{4} + {\left (b^{4} d^{4} e^{4} - 4 \, a b^{3} d^{3} e^{5} + 6 \, a^{2} b^{2} d^{2} e^{6} - 4 \, a^{3} b d e^{7} + a^{4} e^{8}\right )} x^{4} + 4 \, {\left (b^{4} d^{5} e^{3} - 4 \, a b^{3} d^{4} e^{4} + 6 \, a^{2} b^{2} d^{3} e^{5} - 4 \, a^{3} b d^{2} e^{6} + a^{4} d e^{7}\right )} x^{3} + 6 \, {\left (b^{4} d^{6} e^{2} - 4 \, a b^{3} d^{5} e^{3} + 6 \, a^{2} b^{2} d^{4} e^{4} - 4 \, a^{3} b d^{3} e^{5} + a^{4} d^{2} e^{6}\right )} x^{2} + 4 \, {\left (b^{4} d^{7} e - 4 \, a b^{3} d^{6} e^{2} + 6 \, a^{2} b^{2} d^{5} e^{3} - 4 \, a^{3} b d^{4} e^{4} + a^{4} d^{3} e^{5}\right )} x\right )}} \]

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)^(9/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2/105*(1575*b^3*d^5 - 2975*a*b^2*d^4*e + 1953*a^2*b*d^3*e^2 - 409*a^3*d^2*e^3 + 16*(58*b^3*d^2*e^3 - 84*a*b^2*
d*e^4 + 35*a^2*b*e^5)*x^3 + 8*(406*b^3*d^3*e^2 - 646*a*b^2*d^2*e^3 + 329*a^2*b*d*e^4 - 35*a^3*e^5)*x^2 + 2*(19
25*b^3*d^4*e - 3332*a*b^2*d^3*e^2 + 1945*a^2*b*d^2*e^3 - 322*a^3*d*e^4)*x)*sqrt(b*x + a)*sqrt(e*x + d)/(b^4*d^
8 - 4*a*b^3*d^7*e + 6*a^2*b^2*d^6*e^2 - 4*a^3*b*d^5*e^3 + a^4*d^4*e^4 + (b^4*d^4*e^4 - 4*a*b^3*d^3*e^5 + 6*a^2
*b^2*d^2*e^6 - 4*a^3*b*d*e^7 + a^4*e^8)*x^4 + 4*(b^4*d^5*e^3 - 4*a*b^3*d^4*e^4 + 6*a^2*b^2*d^3*e^5 - 4*a^3*b*d
^2*e^6 + a^4*d*e^7)*x^3 + 6*(b^4*d^6*e^2 - 4*a*b^3*d^5*e^3 + 6*a^2*b^2*d^4*e^4 - 4*a^3*b*d^3*e^5 + a^4*d^2*e^6
)*x^2 + 4*(b^4*d^7*e - 4*a*b^3*d^6*e^2 + 6*a^2*b^2*d^5*e^3 - 4*a^3*b*d^4*e^4 + a^4*d^3*e^5)*x)

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

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)^(9/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 248, normalized size = 1.31 \[ -\frac {2 \sqrt {b x +a}\, \left (-560 a^{2} b \,e^{5} x^{3}+1344 a \,b^{2} d \,e^{4} x^{3}-928 b^{3} d^{2} e^{3} x^{3}+280 a^{3} e^{5} x^{2}-2632 a^{2} b d \,e^{4} x^{2}+5168 a \,b^{2} d^{2} e^{3} x^{2}-3248 b^{3} d^{3} e^{2} x^{2}+644 a^{3} d \,e^{4} x -3890 a^{2} b \,d^{2} e^{3} x +6664 a \,b^{2} d^{3} e^{2} x -3850 b^{3} d^{4} e x +409 a^{3} d^{2} e^{3}-1953 a^{2} b \,d^{3} e^{2}+2975 a \,b^{2} d^{4} e -1575 b^{3} d^{5}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} \left (e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )} \]

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)^(9/2)/(b*x+a)^(1/2),x)

[Out]

-2/105*(b*x+a)^(1/2)*(-560*a^2*b*e^5*x^3+1344*a*b^2*d*e^4*x^3-928*b^3*d^2*e^3*x^3+280*a^3*e^5*x^2-2632*a^2*b*d
*e^4*x^2+5168*a*b^2*d^2*e^3*x^2-3248*b^3*d^3*e^2*x^2+644*a^3*d*e^4*x-3890*a^2*b*d^2*e^3*x+6664*a*b^2*d^3*e^2*x
-3850*b^3*d^4*e*x+409*a^3*d^2*e^3-1953*a^2*b*d^3*e^2+2975*a*b^2*d^4*e-1575*b^3*d^5)/(e*x+d)^(7/2)/(a^4*e^4-4*a
^3*b*d*e^3+6*a^2*b^2*d^2*e^2-4*a*b^3*d^3*e+b^4*d^4)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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)^(9/2)/(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 [B]  time = 4.51, size = 389, normalized size = 2.06 \[ \frac {\sqrt {d+e\,x}\,\left (\frac {-818\,a^4\,d^2\,e^3+3906\,a^3\,b\,d^3\,e^2-5950\,a^2\,b^2\,d^4\,e+3150\,a\,b^3\,d^5}{105\,e^4\,{\left (a\,e-b\,d\right )}^4}+\frac {x\,\left (-1288\,a^4\,d\,e^4+6962\,a^3\,b\,d^2\,e^3-9422\,a^2\,b^2\,d^3\,e^2+1750\,a\,b^3\,d^4\,e+3150\,b^4\,d^5\right )}{105\,e^4\,{\left (a\,e-b\,d\right )}^4}-\frac {x^2\,\left (560\,a^4\,e^5-3976\,a^3\,b\,d\,e^4+2556\,a^2\,b^2\,d^2\,e^3+6832\,a\,b^3\,d^3\,e^2-7700\,b^4\,d^4\,e\right )}{105\,e^4\,{\left (a\,e-b\,d\right )}^4}+\frac {32\,b^2\,x^4\,\left (35\,a^2\,e^2-84\,a\,b\,d\,e+58\,b^2\,d^2\right )}{105\,e\,{\left (a\,e-b\,d\right )}^4}+\frac {16\,b\,x^3\,\left (35\,a^3\,e^3+161\,a^2\,b\,d\,e^2-530\,a\,b^2\,d^2\,e+406\,b^3\,d^3\right )}{105\,e^2\,{\left (a\,e-b\,d\right )}^4}\right )}{x^4\,\sqrt {a+b\,x}+\frac {d^4\,\sqrt {a+b\,x}}{e^4}+\frac {6\,d^2\,x^2\,\sqrt {a+b\,x}}{e^2}+\frac {4\,d\,x^3\,\sqrt {a+b\,x}}{e}+\frac {4\,d^3\,x\,\sqrt {a+b\,x}}{e^3}} \]

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)^(9/2)),x)

[Out]

((d + e*x)^(1/2)*((3150*a*b^3*d^5 - 818*a^4*d^2*e^3 - 5950*a^2*b^2*d^4*e + 3906*a^3*b*d^3*e^2)/(105*e^4*(a*e -
b*d)^4) + (x*(3150*b^4*d^5 - 1288*a^4*d*e^4 + 6962*a^3*b*d^2*e^3 - 9422*a^2*b^2*d^3*e^2 + 1750*a*b^3*d^4*e))/
(105*e^4*(a*e - b*d)^4) - (x^2*(560*a^4*e^5 - 7700*b^4*d^4*e + 6832*a*b^3*d^3*e^2 + 2556*a^2*b^2*d^2*e^3 - 397
6*a^3*b*d*e^4))/(105*e^4*(a*e - b*d)^4) + (32*b^2*x^4*(35*a^2*e^2 + 58*b^2*d^2 - 84*a*b*d*e))/(105*e*(a*e - b*
d)^4) + (16*b*x^3*(35*a^3*e^3 + 406*b^3*d^3 - 530*a*b^2*d^2*e + 161*a^2*b*d*e^2))/(105*e^2*(a*e - b*d)^4)))/(x
^4*(a + b*x)^(1/2) + (d^4*(a + b*x)^(1/2))/e^4 + (6*d^2*x^2*(a + b*x)^(1/2))/e^2 + (4*d*x^3*(a + b*x)^(1/2))/e
+ (4*d^3*x*(a + b*x)^(1/2))/e^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(9/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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