3.18.57 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^7} \, dx\)

Optimal. Leaf size=98 \[ \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{30 (d+e x)^5 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{6 (d+e x)^6 (b d-a e)} \]

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Rubi [A]  time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, 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 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{30 (d+e x)^5 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{6 (d+e x)^6 (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)^7,x]

[Out]

((a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*(b*d - a*e)*(d + e*x)^6) + (b*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(30*(b*d - a*e)^2*(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)^7} \, 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)^7} \, 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)^7} \, dx}{a b+b^2 x}\\ &=\frac {(a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (b d-a e) (d+e x)^6}+\frac {\left (b^2 \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^4}{(d+e x)^6} \, dx}{6 (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}}{6 (b d-a e) (d+e x)^6}+\frac {b (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{30 (b d-a e)^2 (d+e x)^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 162, normalized size = 1.65 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (5 a^4 e^4+4 a^3 b e^3 (d+6 e x)+3 a^2 b^2 e^2 \left (d^2+6 d e x+15 e^2 x^2\right )+2 a b^3 e \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )+b^4 \left (d^4+6 d^3 e x+15 d^2 e^2 x^2+20 d e^3 x^3+15 e^4 x^4\right )\right )}{30 e^5 (a+b x) (d+e x)^6} \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)^7,x]

[Out]

-1/30*(Sqrt[(a + b*x)^2]*(5*a^4*e^4 + 4*a^3*b*e^3*(d + 6*e*x) + 3*a^2*b^2*e^2*(d^2 + 6*d*e*x + 15*e^2*x^2) + 2
*a*b^3*e*(d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3) + b^4*(d^4 + 6*d^3*e*x + 15*d^2*e^2*x^2 + 20*d*e^3*x^3
+ 15*e^4*x^4)))/(e^5*(a + b*x)*(d + e*x)^6)

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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)^7,x]

[Out]

$Aborted

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fricas [B]  time = 0.42, size = 236, normalized size = 2.41 \begin {gather*} -\frac {15 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + 2 \, a b^{3} d^{3} e + 3 \, a^{2} b^{2} d^{2} e^{2} + 4 \, a^{3} b d e^{3} + 5 \, a^{4} e^{4} + 20 \, {\left (b^{4} d e^{3} + 2 \, a b^{3} e^{4}\right )} x^{3} + 15 \, {\left (b^{4} d^{2} e^{2} + 2 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} + 6 \, {\left (b^{4} d^{3} e + 2 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} + 4 \, a^{3} b e^{4}\right )} x}{30 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} 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)^7,x, algorithm="fricas")

[Out]

-1/30*(15*b^4*e^4*x^4 + b^4*d^4 + 2*a*b^3*d^3*e + 3*a^2*b^2*d^2*e^2 + 4*a^3*b*d*e^3 + 5*a^4*e^4 + 20*(b^4*d*e^
3 + 2*a*b^3*e^4)*x^3 + 15*(b^4*d^2*e^2 + 2*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 + 6*(b^4*d^3*e + 2*a*b^3*d^2*e^2 +
3*a^2*b^2*d*e^3 + 4*a^3*b*e^4)*x)/(e^11*x^6 + 6*d*e^10*x^5 + 15*d^2*e^9*x^4 + 20*d^3*e^8*x^3 + 15*d^4*e^7*x^2
+ 6*d^5*e^6*x + d^6*e^5)

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giac [B]  time = 0.21, size = 264, normalized size = 2.69 \begin {gather*} -\frac {{\left (15 \, b^{4} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 20 \, b^{4} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, b^{4} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{4} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) + 40 \, a b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 30 \, a b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 12 \, a b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 45 \, a^{2} b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 24 \, a^{3} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{30 \, {\left (x e + d\right )}^{6}} \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)^7,x, algorithm="giac")

[Out]

-1/30*(15*b^4*x^4*e^4*sgn(b*x + a) + 20*b^4*d*x^3*e^3*sgn(b*x + a) + 15*b^4*d^2*x^2*e^2*sgn(b*x + a) + 6*b^4*d
^3*x*e*sgn(b*x + a) + b^4*d^4*sgn(b*x + a) + 40*a*b^3*x^3*e^4*sgn(b*x + a) + 30*a*b^3*d*x^2*e^3*sgn(b*x + a) +
12*a*b^3*d^2*x*e^2*sgn(b*x + a) + 2*a*b^3*d^3*e*sgn(b*x + a) + 45*a^2*b^2*x^2*e^4*sgn(b*x + a) + 18*a^2*b^2*d
*x*e^3*sgn(b*x + a) + 3*a^2*b^2*d^2*e^2*sgn(b*x + a) + 24*a^3*b*x*e^4*sgn(b*x + a) + 4*a^3*b*d*e^3*sgn(b*x + a
) + 5*a^4*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^6

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maple [B]  time = 0.05, size = 201, normalized size = 2.05 \begin {gather*} -\frac {\left (15 b^{4} e^{4} x^{4}+40 a \,b^{3} e^{4} x^{3}+20 b^{4} d \,e^{3} x^{3}+45 a^{2} b^{2} e^{4} x^{2}+30 a \,b^{3} d \,e^{3} x^{2}+15 b^{4} d^{2} e^{2} x^{2}+24 a^{3} b \,e^{4} x +18 a^{2} b^{2} d \,e^{3} x +12 a \,b^{3} d^{2} e^{2} x +6 b^{4} d^{3} e x +5 a^{4} e^{4}+4 a^{3} b d \,e^{3}+3 a^{2} b^{2} d^{2} e^{2}+2 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{30 \left (e x +d \right )^{6} \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)^7,x)

[Out]

-1/30/e^5*(15*b^4*e^4*x^4+40*a*b^3*e^4*x^3+20*b^4*d*e^3*x^3+45*a^2*b^2*e^4*x^2+30*a*b^3*d*e^3*x^2+15*b^4*d^2*e
^2*x^2+24*a^3*b*e^4*x+18*a^2*b^2*d*e^3*x+12*a*b^3*d^2*e^2*x+6*b^4*d^3*e*x+5*a^4*e^4+4*a^3*b*d*e^3+3*a^2*b^2*d^
2*e^2+2*a*b^3*d^3*e+b^4*d^4)*((b*x+a)^2)^(3/2)/(e*x+d)^6/(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)^7,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 = 2.14, size = 449, normalized size = 4.58 \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}{5\,e^5}+\frac {d\,\left (\frac {d\,\left (\frac {b^4\,d}{5\,e^3}-\frac {b^3\,\left (4\,a\,e-b\,d\right )}{5\,e^3}\right )}{e}+\frac {b^2\,\left (6\,a^2\,e^2-4\,a\,b\,d\,e+b^2\,d^2\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 {a^4}{6\,e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {2\,a\,b^3}{3\,e}-\frac {b^4\,d}{6\,e^2}\right )}{e}-\frac {a^2\,b^2}{e}\right )}{e}+\frac {2\,a^3\,b}{3\,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 )}^6}-\frac {\left (\frac {6\,a^2\,b^2\,e^2-8\,a\,b^3\,d\,e+3\,b^4\,d^2}{4\,e^5}+\frac {d\,\left (\frac {b^4\,d}{4\,e^4}-\frac {b^3\,\left (2\,a\,e-b\,d\right )}{2\,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 )}^4}+\frac {\left (\frac {3\,b^4\,d-4\,a\,b^3\,e}{3\,e^5}+\frac {b^4\,d}{3\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}-\frac {b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2} \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)^7,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)/(5*e^5) + (d*((d*((b^4*d)/(5*e^3) - (b^3*(4*a*e -
b*d))/(5*e^3)))/e + (b^2*(6*a^2*e^2 + b^2*d^2 - 4*a*b*d*e))/(5*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a
+ b*x)*(d + e*x)^5) - ((a^4/(6*e) - (d*((d*((d*((2*a*b^3)/(3*e) - (b^4*d)/(6*e^2)))/e - (a^2*b^2)/e))/e + (2*a
^3*b)/(3*e)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^6) - (((3*b^4*d^2 + 6*a^2*b^2*e^2 - 8*a
*b^3*d*e)/(4*e^5) + (d*((b^4*d)/(4*e^4) - (b^3*(2*a*e - b*d))/(2*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((
a + b*x)*(d + e*x)^4) + (((3*b^4*d - 4*a*b^3*e)/(3*e^5) + (b^4*d)/(3*e^5))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((
a + b*x)*(d + e*x)^3) - (b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*e^5*(a + b*x)*(d + e*x)^2)

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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 )^{7}}\, 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)**7,x)

[Out]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**7, x)

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