### 3.1566 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^7} \, dx$$

Optimal. Leaf size=143 $\frac{b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{60 (d+e x)^4 (b d-a e)^3}+\frac{b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{15 (d+e x)^5 (b d-a e)^2}+\frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{6 (d+e x)^6 (b d-a e)}$

[Out]

((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(6*(b*d - a*e)*(d + e*x)^6) + (b*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^
2)^(3/2))/(15*(b*d - a*e)^2*(d + e*x)^5) + (b^2*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(60*(b*d - a*e)^3*(
d + e*x)^4)

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Rubi [A]  time = 0.0861064, antiderivative size = 200, normalized size of antiderivative = 1.4, number of steps used = 3, number of rules used = 2, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {646, 43} $-\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x) (d+e x)^3}+\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{4 e^4 (a+b x) (d+e x)^4}-\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^4 (a+b x) (d+e x)^5}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{6 e^4 (a+b x) (d+e x)^6}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^7,x]

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0435135, size = 112, normalized size = 0.78 $-\frac{\sqrt{(a+b x)^2} \left (6 a^2 b e^2 (d+6 e x)+10 a^3 e^3+3 a b^2 e \left (d^2+6 d e x+15 e^2 x^2\right )+b^3 \left (6 d^2 e x+d^3+15 d e^2 x^2+20 e^3 x^3\right )\right )}{60 e^4 (a+b x) (d+e x)^6}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^7,x]

[Out]

-(Sqrt[(a + b*x)^2]*(10*a^3*e^3 + 6*a^2*b*e^2*(d + 6*e*x) + 3*a*b^2*e*(d^2 + 6*d*e*x + 15*e^2*x^2) + b^3*(d^3
+ 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3)))/(60*e^4*(a + b*x)*(d + e*x)^6)

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Maple [A]  time = 0.159, size = 131, normalized size = 0.9 \begin{align*} -{\frac{20\,{x}^{3}{b}^{3}{e}^{3}+45\,{x}^{2}a{b}^{2}{e}^{3}+15\,{x}^{2}{b}^{3}d{e}^{2}+36\,x{a}^{2}b{e}^{3}+18\,xa{b}^{2}d{e}^{2}+6\,x{b}^{3}{d}^{2}e+10\,{a}^{3}{e}^{3}+6\,d{e}^{2}{a}^{2}b+3\,a{b}^{2}{d}^{2}e+{b}^{3}{d}^{3}}{60\,{e}^{4} \left ( ex+d \right ) ^{6} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/e^4*(20*b^3*e^3*x^3+45*a*b^2*e^3*x^2+15*b^3*d*e^2*x^2+36*a^2*b*e^3*x+18*a*b^2*d*e^2*x+6*b^3*d^2*e*x+10*a
^3*e^3+6*a^2*b*d*e^2+3*a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^6/(b*x+a)^3

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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((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84614, size = 354, normalized size = 2.48 \begin{align*} -\frac{20 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 3 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 10 \, a^{3} e^{3} + 15 \,{\left (b^{3} d e^{2} + 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \,{\left (b^{3} d^{2} e + 3 \, a b^{2} d e^{2} + 6 \, a^{2} b e^{3}\right )} x}{60 \,{\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^7,x, algorithm="fricas")

[Out]

-1/60*(20*b^3*e^3*x^3 + b^3*d^3 + 3*a*b^2*d^2*e + 6*a^2*b*d*e^2 + 10*a^3*e^3 + 15*(b^3*d*e^2 + 3*a*b^2*e^3)*x^
2 + 6*(b^3*d^2*e + 3*a*b^2*d*e^2 + 6*a^2*b*e^3)*x)/(e^10*x^6 + 6*d*e^9*x^5 + 15*d^2*e^8*x^4 + 20*d^3*e^7*x^3 +
15*d^4*e^6*x^2 + 6*d^5*e^5*x + d^6*e^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.14024, size = 228, normalized size = 1.59 \begin{align*} -\frac{{\left (20 \, b^{3} x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 15 \, b^{3} d x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, b^{3} d^{2} x e \mathrm{sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) + 45 \, a b^{2} x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 18 \, a b^{2} d x e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 36 \, a^{2} b x e^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) + 10 \, a^{3} e^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{60 \,{\left (x e + d\right )}^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^7,x, algorithm="giac")

[Out]

-1/60*(20*b^3*x^3*e^3*sgn(b*x + a) + 15*b^3*d*x^2*e^2*sgn(b*x + a) + 6*b^3*d^2*x*e*sgn(b*x + a) + b^3*d^3*sgn(
b*x + a) + 45*a*b^2*x^2*e^3*sgn(b*x + a) + 18*a*b^2*d*x*e^2*sgn(b*x + a) + 3*a*b^2*d^2*e*sgn(b*x + a) + 36*a^2
*b*x*e^3*sgn(b*x + a) + 6*a^2*b*d*e^2*sgn(b*x + a) + 10*a^3*e^3*sgn(b*x + a))*e^(-4)/(x*e + d)^6