3.18.58 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^8} \, dx\)

Optimal. Leaf size=149 \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{105 (d+e x)^5 (b d-a e)^3}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{21 (d+e x)^6 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{7 (d+e x)^7 (b d-a e)} \]

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Rubi [A]  time = 0.07, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {770, 21, 45, 37} \begin {gather*} \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{105 (d+e x)^5 (b d-a e)^3}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{21 (d+e x)^6 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{7 (d+e x)^7 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^8,x]

[Out]

((a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(b*d - a*e)*(d + e*x)^7) + (b*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(21*(b*d - a*e)^2*(d + e*x)^6) + (b^2*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(105*(b*d - a*e)^3*
(d + e*x)^5)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) \left (a b+b^2 x\right )^3}{(d+e x)^8} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^4}{(d+e x)^8} \, dx}{a b+b^2 x}\\ &=\frac {(a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (b d-a e) (d+e x)^7}+\frac {\left (2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^4}{(d+e x)^7} \, dx}{7 (b d-a e) \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (b d-a e) (d+e x)^7}+\frac {b (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{21 (b d-a e)^2 (d+e x)^6}+\frac {\left (b^3 \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^4}{(d+e x)^6} \, dx}{21 (b d-a e)^2 \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (b d-a e) (d+e x)^7}+\frac {b (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{21 (b d-a e)^2 (d+e x)^6}+\frac {b^2 (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{105 (b d-a e)^3 (d+e x)^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 162, normalized size = 1.09 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (15 a^4 e^4+10 a^3 b e^3 (d+7 e x)+6 a^2 b^2 e^2 \left (d^2+7 d e x+21 e^2 x^2\right )+3 a b^3 e \left (d^3+7 d^2 e x+21 d e^2 x^2+35 e^3 x^3\right )+b^4 \left (d^4+7 d^3 e x+21 d^2 e^2 x^2+35 d e^3 x^3+35 e^4 x^4\right )\right )}{105 e^5 (a+b x) (d+e x)^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^8,x]

[Out]

-1/105*(Sqrt[(a + b*x)^2]*(15*a^4*e^4 + 10*a^3*b*e^3*(d + 7*e*x) + 6*a^2*b^2*e^2*(d^2 + 7*d*e*x + 21*e^2*x^2)
+ 3*a*b^3*e*(d^3 + 7*d^2*e*x + 21*d*e^2*x^2 + 35*e^3*x^3) + b^4*(d^4 + 7*d^3*e*x + 21*d^2*e^2*x^2 + 35*d*e^3*x
^3 + 35*e^4*x^4)))/(e^5*(a + b*x)*(d + e*x)^7)

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IntegrateAlgebraic [F]  time = 180.03, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^8,x]

[Out]

$Aborted

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fricas [B]  time = 0.45, size = 247, normalized size = 1.66 \begin {gather*} -\frac {35 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + 3 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} + 10 \, a^{3} b d e^{3} + 15 \, a^{4} e^{4} + 35 \, {\left (b^{4} d e^{3} + 3 \, a b^{3} e^{4}\right )} x^{3} + 21 \, {\left (b^{4} d^{2} e^{2} + 3 \, a b^{3} d e^{3} + 6 \, a^{2} b^{2} e^{4}\right )} x^{2} + 7 \, {\left (b^{4} d^{3} e + 3 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + 10 \, a^{3} b e^{4}\right )} x}{105 \, {\left (e^{12} x^{7} + 7 \, d e^{11} x^{6} + 21 \, d^{2} e^{10} x^{5} + 35 \, d^{3} e^{9} x^{4} + 35 \, d^{4} e^{8} x^{3} + 21 \, d^{5} e^{7} x^{2} + 7 \, d^{6} e^{6} x + d^{7} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x, algorithm="fricas")

[Out]

-1/105*(35*b^4*e^4*x^4 + b^4*d^4 + 3*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 10*a^3*b*d*e^3 + 15*a^4*e^4 + 35*(b^4*d
*e^3 + 3*a*b^3*e^4)*x^3 + 21*(b^4*d^2*e^2 + 3*a*b^3*d*e^3 + 6*a^2*b^2*e^4)*x^2 + 7*(b^4*d^3*e + 3*a*b^3*d^2*e^
2 + 6*a^2*b^2*d*e^3 + 10*a^3*b*e^4)*x)/(e^12*x^7 + 7*d*e^11*x^6 + 21*d^2*e^10*x^5 + 35*d^3*e^9*x^4 + 35*d^4*e^
8*x^3 + 21*d^5*e^7*x^2 + 7*d^6*e^6*x + d^7*e^5)

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giac [B]  time = 0.17, size = 264, normalized size = 1.77 \begin {gather*} -\frac {{\left (35 \, b^{4} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 35 \, b^{4} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 21 \, b^{4} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 7 \, b^{4} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) + 105 \, a b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 63 \, a b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 21 \, a b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 126 \, a^{2} b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 42 \, a^{2} b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 70 \, a^{3} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{105 \, {\left (x e + d\right )}^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x, algorithm="giac")

[Out]

-1/105*(35*b^4*x^4*e^4*sgn(b*x + a) + 35*b^4*d*x^3*e^3*sgn(b*x + a) + 21*b^4*d^2*x^2*e^2*sgn(b*x + a) + 7*b^4*
d^3*x*e*sgn(b*x + a) + b^4*d^4*sgn(b*x + a) + 105*a*b^3*x^3*e^4*sgn(b*x + a) + 63*a*b^3*d*x^2*e^3*sgn(b*x + a)
+ 21*a*b^3*d^2*x*e^2*sgn(b*x + a) + 3*a*b^3*d^3*e*sgn(b*x + a) + 126*a^2*b^2*x^2*e^4*sgn(b*x + a) + 42*a^2*b^
2*d*x*e^3*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) + 70*a^3*b*x*e^4*sgn(b*x + a) + 10*a^3*b*d*e^3*sgn(b*x
+ a) + 15*a^4*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^7

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maple [A]  time = 0.05, size = 201, normalized size = 1.35 \begin {gather*} -\frac {\left (35 b^{4} e^{4} x^{4}+105 a \,b^{3} e^{4} x^{3}+35 b^{4} d \,e^{3} x^{3}+126 a^{2} b^{2} e^{4} x^{2}+63 a \,b^{3} d \,e^{3} x^{2}+21 b^{4} d^{2} e^{2} x^{2}+70 a^{3} b \,e^{4} x +42 a^{2} b^{2} d \,e^{3} x +21 a \,b^{3} d^{2} e^{2} x +7 b^{4} d^{3} e x +15 a^{4} e^{4}+10 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}+3 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{105 \left (e x +d \right )^{7} \left (b x +a \right )^{3} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x)

[Out]

-1/105/e^5*(35*b^4*e^4*x^4+105*a*b^3*e^4*x^3+35*b^4*d*e^3*x^3+126*a^2*b^2*e^4*x^2+63*a*b^3*d*e^3*x^2+21*b^4*d^
2*e^2*x^2+70*a^3*b*e^4*x+42*a^2*b^2*d*e^3*x+21*a*b^3*d^2*e^2*x+7*b^4*d^3*e*x+15*a^4*e^4+10*a^3*b*d*e^3+6*a^2*b
^2*d^2*e^2+3*a*b^3*d^3*e+b^4*d^4)*((b*x+a)^2)^(3/2)/(e*x+d)^7/(b*x+a)^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((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,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 = 2.15, size = 449, normalized size = 3.01 \begin {gather*} \frac {\left (\frac {-4\,a^3\,b\,e^3+6\,a^2\,b^2\,d\,e^2-4\,a\,b^3\,d^2\,e+b^4\,d^3}{6\,e^5}+\frac {d\,\left (\frac {d\,\left (\frac {b^4\,d}{6\,e^3}-\frac {b^3\,\left (4\,a\,e-b\,d\right )}{6\,e^3}\right )}{e}+\frac {b^2\,\left (6\,a^2\,e^2-4\,a\,b\,d\,e+b^2\,d^2\right )}{6\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {a^4}{7\,e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {4\,a\,b^3}{7\,e}-\frac {b^4\,d}{7\,e^2}\right )}{e}-\frac {6\,a^2\,b^2}{7\,e}\right )}{e}+\frac {4\,a^3\,b}{7\,e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {6\,a^2\,b^2\,e^2-8\,a\,b^3\,d\,e+3\,b^4\,d^2}{5\,e^5}+\frac {d\,\left (\frac {b^4\,d}{5\,e^4}-\frac {2\,b^3\,\left (2\,a\,e-b\,d\right )}{5\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}+\frac {\left (\frac {3\,b^4\,d-4\,a\,b^3\,e}{4\,e^5}+\frac {b^4\,d}{4\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^8,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{8}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**8,x)

[Out]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**8, x)

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