3.9.48 \(\int \frac {15 d^2+20 d e x+8 e^2 x^2}{\sqrt {a+b x} (d+e x)^{7/2}} \, dx\) [848]

Optimal. Leaf size=133 \[ \frac {6 d^2 \sqrt {a+b x}}{5 (b d-a e) (d+e x)^{5/2}}+\frac {8 d (8 b d-5 a e) \sqrt {a+b x}}{15 (b d-a e)^2 (d+e x)^{3/2}}+\frac {16 \left (23 b^2 d^2-35 a b d e+15 a^2 e^2\right ) \sqrt {a+b x}}{15 (b d-a e)^3 \sqrt {d+e x}} \]

[Out]

6/5*d^2*(b*x+a)^(1/2)/(-a*e+b*d)/(e*x+d)^(5/2)+8/15*d*(-5*a*e+8*b*d)*(b*x+a)^(1/2)/(-a*e+b*d)^2/(e*x+d)^(3/2)+
16/15*(15*a^2*e^2-35*a*b*d*e+23*b^2*d^2)*(b*x+a)^(1/2)/(-a*e+b*d)^3/(e*x+d)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {963, 79, 37} \begin {gather*} \frac {16 \sqrt {a+b x} \left (15 a^2 e^2-35 a b d e+23 b^2 d^2\right )}{15 \sqrt {d+e x} (b d-a e)^3}+\frac {6 d^2 \sqrt {a+b x}}{5 (d+e x)^{5/2} (b d-a e)}+\frac {8 d \sqrt {a+b x} (8 b d-5 a e)}{15 (d+e x)^{3/2} (b d-a e)^2} \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)^(7/2)),x]

[Out]

(6*d^2*Sqrt[a + b*x])/(5*(b*d - a*e)*(d + e*x)^(5/2)) + (8*d*(8*b*d - 5*a*e)*Sqrt[a + b*x])/(15*(b*d - a*e)^2*
(d + e*x)^(3/2)) + (16*(23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2)*Sqrt[a + b*x])/(15*(b*d - a*e)^3*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 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 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)^{7/2}} \, dx &=\frac {6 d^2 \sqrt {a+b x}}{5 (b d-a e) (d+e x)^{5/2}}+\frac {2 \int \frac {6 d (6 b d-5 a e)+20 e (b d-a e) x}{\sqrt {a+b x} (d+e x)^{5/2}} \, dx}{5 (b d-a e)}\\ &=\frac {6 d^2 \sqrt {a+b x}}{5 (b d-a e) (d+e x)^{5/2}}+\frac {8 d (8 b d-5 a e) \sqrt {a+b x}}{15 (b d-a e)^2 (d+e x)^{3/2}}+\frac {\left (8 \left (23 b^2 d^2-35 a b d e+15 a^2 e^2\right )\right ) \int \frac {1}{\sqrt {a+b x} (d+e x)^{3/2}} \, dx}{15 (b d-a e)^2}\\ &=\frac {6 d^2 \sqrt {a+b x}}{5 (b d-a e) (d+e x)^{5/2}}+\frac {8 d (8 b d-5 a e) \sqrt {a+b x}}{15 (b d-a e)^2 (d+e x)^{3/2}}+\frac {16 \left (23 b^2 d^2-35 a b d e+15 a^2 e^2\right ) \sqrt {a+b x}}{15 (b d-a e)^3 \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 115, normalized size = 0.86 \begin {gather*} \frac {2 \sqrt {a+b x} \left (225 b^2 d^2-300 a b d e+120 a^2 e^2+\frac {9 d^2 e^2 (a+b x)^2}{(d+e x)^2}-\frac {50 b d^2 e (a+b x)}{d+e x}+\frac {20 a d e^2 (a+b x)}{d+e x}\right )}{15 (b d-a e)^3 \sqrt {d+e x}} \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)^(7/2)),x]

[Out]

(2*Sqrt[a + b*x]*(225*b^2*d^2 - 300*a*b*d*e + 120*a^2*e^2 + (9*d^2*e^2*(a + b*x)^2)/(d + e*x)^2 - (50*b*d^2*e*
(a + b*x))/(d + e*x) + (20*a*d*e^2*(a + b*x))/(d + e*x)))/(15*(b*d - a*e)^3*Sqrt[d + e*x])

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Maple [A]
time = 0.08, size = 122, normalized size = 0.92

method	result	size
default	\(-\frac {2 \sqrt {b x +a}\, \left (120 a^{2} e^{4} x^{2}-280 a b d \,e^{3} x^{2}+184 b^{2} d^{2} e^{2} x^{2}+260 a^{2} d \,e^{3} x -612 a b \,d^{2} e^{2} x +400 b^{2} d^{3} e x +149 a^{2} d^{2} e^{2}-350 a b \,d^{3} e +225 b^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} \left (a e -b d \right )^{3}}\)	\(122\)
gosper	\(-\frac {2 \sqrt {b x +a}\, \left (120 a^{2} e^{4} x^{2}-280 a b d \,e^{3} x^{2}+184 b^{2} d^{2} e^{2} x^{2}+260 a^{2} d \,e^{3} x -612 a b \,d^{2} e^{2} x +400 b^{2} d^{3} e x +149 a^{2} d^{2} e^{2}-350 a b \,d^{3} e +225 b^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\)	\(150\)

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)^(7/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*(b*x+a)^(1/2)*(120*a^2*e^4*x^2-280*a*b*d*e^3*x^2+184*b^2*d^2*e^2*x^2+260*a^2*d*e^3*x-612*a*b*d^2*e^2*x+4
00*b^2*d^3*e*x+149*a^2*d^2*e^2-350*a*b*d^3*e+225*b^2*d^4)/(e*x+d)^(5/2)/(a*e-b*d)^3

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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)^(7/2)/(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. 286 vs. \(2 (122) = 244\).
time = 5.38, size = 286, normalized size = 2.15 \begin {gather*} \frac {2 \, {\left (225 \, b^{2} d^{4} + 120 \, a^{2} x^{2} e^{4} - 20 \, {\left (14 \, a b d x^{2} - 13 \, a^{2} d x\right )} e^{3} + {\left (184 \, b^{2} d^{2} x^{2} - 612 \, a b d^{2} x + 149 \, a^{2} d^{2}\right )} e^{2} + 50 \, {\left (8 \, b^{2} d^{3} x - 7 \, a b d^{3}\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{15 \, {\left (b^{3} d^{6} - a^{3} x^{3} e^{6} + 3 \, {\left (a^{2} b d x^{3} - a^{3} d x^{2}\right )} e^{5} - 3 \, {\left (a b^{2} d^{2} x^{3} - 3 \, a^{2} b d^{2} x^{2} + a^{3} d^{2} x\right )} e^{4} + {\left (b^{3} d^{3} x^{3} - 9 \, a b^{2} d^{3} x^{2} + 9 \, a^{2} b d^{3} x - a^{3} d^{3}\right )} e^{3} + 3 \, {\left (b^{3} d^{4} x^{2} - 3 \, a b^{2} d^{4} x + a^{2} b d^{4}\right )} e^{2} + 3 \, {\left (b^{3} d^{5} x - a b^{2} d^{5}\right )} e\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)^(7/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2/15*(225*b^2*d^4 + 120*a^2*x^2*e^4 - 20*(14*a*b*d*x^2 - 13*a^2*d*x)*e^3 + (184*b^2*d^2*x^2 - 612*a*b*d^2*x +
149*a^2*d^2)*e^2 + 50*(8*b^2*d^3*x - 7*a*b*d^3)*e)*sqrt(b*x + a)*sqrt(x*e + d)/(b^3*d^6 - a^3*x^3*e^6 + 3*(a^2
*b*d*x^3 - a^3*d*x^2)*e^5 - 3*(a*b^2*d^2*x^3 - 3*a^2*b*d^2*x^2 + a^3*d^2*x)*e^4 + (b^3*d^3*x^3 - 9*a*b^2*d^3*x
^2 + 9*a^2*b*d^3*x - a^3*d^3)*e^3 + 3*(b^3*d^4*x^2 - 3*a*b^2*d^4*x + a^2*b*d^4)*e^2 + 3*(b^3*d^5*x - a*b^2*d^5
)*e)

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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 {7}{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)**(7/2)/(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)**(7/2)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (122) = 244\).
time = 3.55, size = 337, normalized size = 2.53 \begin {gather*} \frac {2 \, {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (23 \, b^{8} d^{2} e^{4} - 35 \, a b^{7} d e^{5} + 15 \, a^{2} b^{6} e^{6}\right )} {\left (b x + a\right )}}{b^{5} d^{3} {\left | b \right |} e^{2} - 3 \, a b^{4} d^{2} {\left | b \right |} e^{3} + 3 \, a^{2} b^{3} d {\left | b \right |} e^{4} - a^{3} b^{2} {\left | b \right |} e^{5}} + \frac {5 \, {\left (20 \, b^{9} d^{3} e^{3} - 49 \, a b^{8} d^{2} e^{4} + 41 \, a^{2} b^{7} d e^{5} - 12 \, a^{3} b^{6} e^{6}\right )}}{b^{5} d^{3} {\left | b \right |} e^{2} - 3 \, a b^{4} d^{2} {\left | b \right |} e^{3} + 3 \, a^{2} b^{3} d {\left | b \right |} e^{4} - a^{3} b^{2} {\left | b \right |} e^{5}}\right )} + \frac {15 \, {\left (15 \, b^{10} d^{4} e^{2} - 50 \, a b^{9} d^{3} e^{3} + 63 \, a^{2} b^{8} d^{2} e^{4} - 36 \, a^{3} b^{7} d e^{5} + 8 \, a^{4} b^{6} e^{6}\right )}}{b^{5} d^{3} {\left | b \right |} e^{2} - 3 \, a b^{4} d^{2} {\left | b \right |} e^{3} + 3 \, a^{2} b^{3} d {\left | b \right |} e^{4} - a^{3} b^{2} {\left | b \right |} e^{5}}\right )} \sqrt {b x + a}}{15 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {5}{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)^(7/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2/15*(4*(b*x + a)*(2*(23*b^8*d^2*e^4 - 35*a*b^7*d*e^5 + 15*a^2*b^6*e^6)*(b*x + a)/(b^5*d^3*abs(b)*e^2 - 3*a*b^
4*d^2*abs(b)*e^3 + 3*a^2*b^3*d*abs(b)*e^4 - a^3*b^2*abs(b)*e^5) + 5*(20*b^9*d^3*e^3 - 49*a*b^8*d^2*e^4 + 41*a^
2*b^7*d*e^5 - 12*a^3*b^6*e^6)/(b^5*d^3*abs(b)*e^2 - 3*a*b^4*d^2*abs(b)*e^3 + 3*a^2*b^3*d*abs(b)*e^4 - a^3*b^2*
abs(b)*e^5)) + 15*(15*b^10*d^4*e^2 - 50*a*b^9*d^3*e^3 + 63*a^2*b^8*d^2*e^4 - 36*a^3*b^7*d*e^5 + 8*a^4*b^6*e^6)
/(b^5*d^3*abs(b)*e^2 - 3*a*b^4*d^2*abs(b)*e^3 + 3*a^2*b^3*d*abs(b)*e^4 - a^3*b^2*abs(b)*e^5))*sqrt(b*x + a)/(b
^2*d + (b*x + a)*b*e - a*b*e)^(5/2)

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Mupad [B]
time = 4.30, size = 268, normalized size = 2.02 \begin {gather*} -\frac {\sqrt {d+e\,x}\,\left (\frac {x^2\,\left (240\,a^3\,e^4-40\,a^2\,b\,d\,e^3-856\,a\,b^2\,d^2\,e^2+800\,b^3\,d^3\,e\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^3}+\frac {x\,\left (520\,a^3\,d\,e^3-926\,a^2\,b\,d^2\,e^2+100\,a\,b^2\,d^3\,e+450\,b^3\,d^4\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^3}+\frac {2\,a\,d^2\,\left (149\,a^2\,e^2-350\,a\,b\,d\,e+225\,b^2\,d^2\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^3}+\frac {16\,b\,x^3\,\left (15\,a^2\,e^2-35\,a\,b\,d\,e+23\,b^2\,d^2\right )}{15\,e\,{\left (a\,e-b\,d\right )}^3}\right )}{x^3\,\sqrt {a+b\,x}+\frac {d^3\,\sqrt {a+b\,x}}{e^3}+\frac {3\,d\,x^2\,\sqrt {a+b\,x}}{e}+\frac {3\,d^2\,x\,\sqrt {a+b\,x}}{e^2}} \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)^(7/2)),x)

[Out]

-((d + e*x)^(1/2)*((x^2*(240*a^3*e^4 + 800*b^3*d^3*e - 856*a*b^2*d^2*e^2 - 40*a^2*b*d*e^3))/(15*e^3*(a*e - b*d
)^3) + (x*(450*b^3*d^4 + 520*a^3*d*e^3 - 926*a^2*b*d^2*e^2 + 100*a*b^2*d^3*e))/(15*e^3*(a*e - b*d)^3) + (2*a*d
^2*(149*a^2*e^2 + 225*b^2*d^2 - 350*a*b*d*e))/(15*e^3*(a*e - b*d)^3) + (16*b*x^3*(15*a^2*e^2 + 23*b^2*d^2 - 35
*a*b*d*e))/(15*e*(a*e - b*d)^3)))/(x^3*(a + b*x)^(1/2) + (d^3*(a + b*x)^(1/2))/e^3 + (3*d*x^2*(a + b*x)^(1/2))
/e + (3*d^2*x*(a + b*x)^(1/2))/e^2)

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