### 3.379 $$\int \frac{(d+e x)^{7/2}}{(b x+c x^2)^3} \, dx$$

Optimal. Leaf size=248 $\frac{\sqrt{d+e x} \left (x (2 c d-b e) \left (b^2 e^2-12 b c d e+12 c^2 d^2\right )+b c d^2 (12 c d-11 b e)\right )}{4 b^4 c \left (b x+c x^2\right )}-\frac{d^{3/2} \left (35 b^2 e^2-84 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5}+\frac{(c d-b e)^{3/2} \left (-b^2 e^2-12 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 c^{3/2}}-\frac{(d+e x)^{5/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}$

[Out]

-((d + e*x)^(5/2)*(b*d + (2*c*d - b*e)*x))/(2*b^2*(b*x + c*x^2)^2) + (Sqrt[d + e*x]*(b*c*d^2*(12*c*d - 11*b*e)
+ (2*c*d - b*e)*(12*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*x))/(4*b^4*c*(b*x + c*x^2)) - (d^(3/2)*(48*c^2*d^2 - 84*b
*c*d*e + 35*b^2*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5) + ((c*d - b*e)^(3/2)*(48*c^2*d^2 - 12*b*c*d*e - b
^2*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*c^(3/2))

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Rubi [A]  time = 0.386262, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {738, 818, 826, 1166, 208} $\frac{\sqrt{d+e x} \left (x (2 c d-b e) \left (b^2 e^2-12 b c d e+12 c^2 d^2\right )+b c d^2 (12 c d-11 b e)\right )}{4 b^4 c \left (b x+c x^2\right )}-\frac{d^{3/2} \left (35 b^2 e^2-84 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5}+\frac{(c d-b e)^{3/2} \left (-b^2 e^2-12 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 c^{3/2}}-\frac{(d+e x)^{5/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((d + e*x)^(5/2)*(b*d + (2*c*d - b*e)*x))/(2*b^2*(b*x + c*x^2)^2) + (Sqrt[d + e*x]*(b*c*d^2*(12*c*d - 11*b*e)
+ (2*c*d - b*e)*(12*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*x))/(4*b^4*c*(b*x + c*x^2)) - (d^(3/2)*(48*c^2*d^2 - 84*b
*c*d*e + 35*b^2*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5) + ((c*d - b*e)^(3/2)*(48*c^2*d^2 - 12*b*c*d*e - b
^2*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*c^(3/2))

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(d+e x)^{7/2}}{\left (b x+c x^2\right )^3} \, dx &=-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}-\frac{\int \frac{(d+e x)^{3/2} \left (\frac{1}{2} d (12 c d-11 b e)+\frac{1}{2} e (2 c d-b e) x\right )}{\left (b x+c x^2\right )^2} \, dx}{2 b^2}\\ &=-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b c d^2 (12 c d-11 b e)+(2 c d-b e) \left (12 c^2 d^2-12 b c d e+b^2 e^2\right ) x\right )}{4 b^4 c \left (b x+c x^2\right )}-\frac{\int \frac{-\frac{1}{4} c d^2 \left (48 c^2 d^2-84 b c d e+35 b^2 e^2\right )-\frac{1}{4} e (2 c d-b e) \left (12 c^2 d^2-12 b c d e-b^2 e^2\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{2 b^4 c}\\ &=-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b c d^2 (12 c d-11 b e)+(2 c d-b e) \left (12 c^2 d^2-12 b c d e+b^2 e^2\right ) x\right )}{4 b^4 c \left (b x+c x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{4} d e (2 c d-b e) \left (12 c^2 d^2-12 b c d e-b^2 e^2\right )-\frac{1}{4} c d^2 e \left (48 c^2 d^2-84 b c d e+35 b^2 e^2\right )-\frac{1}{4} e (2 c d-b e) \left (12 c^2 d^2-12 b c d e-b^2 e^2\right ) x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{b^4 c}\\ &=-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b c d^2 (12 c d-11 b e)+(2 c d-b e) \left (12 c^2 d^2-12 b c d e+b^2 e^2\right ) x\right )}{4 b^4 c \left (b x+c x^2\right )}-\frac{\left ((c d-b e)^2 \left (48 c^2 d^2-12 b c d e-b^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{4 b^5 c}+\frac{\left (c d^2 \left (48 c^2 d^2-84 b c d e+35 b^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{4 b^5}\\ &=-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b c d^2 (12 c d-11 b e)+(2 c d-b e) \left (12 c^2 d^2-12 b c d e+b^2 e^2\right ) x\right )}{4 b^4 c \left (b x+c x^2\right )}-\frac{d^{3/2} \left (48 c^2 d^2-84 b c d e+35 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5}+\frac{(c d-b e)^{3/2} \left (48 c^2 d^2-12 b c d e-b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.501917, size = 263, normalized size = 1.06 $\frac{\frac{b \sqrt{d+e x} \left (b^2 c^2 d x \left (8 d^2-55 d e x+10 e^2 x^2\right )+b^3 c \left (-13 d^2 e x-2 d^3+16 d e^2 x^2+e^3 x^3\right )+b^4 \left (-e^3\right ) x^2+36 b c^3 d^2 x^2 (d-e x)+24 c^4 d^3 x^3\right )}{c x^2 (b+c x)^2}+d^{3/2} \left (-\left (35 b^2 e^2-84 b c d e+48 c^2 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )+\frac{\sqrt{c d-b e} \left (11 b^2 c d e^2+b^3 e^3-60 b c^2 d^2 e+48 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{c^{3/2}}}{4 b^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((b*Sqrt[d + e*x]*(-(b^4*e^3*x^2) + 24*c^4*d^3*x^3 + 36*b*c^3*d^2*x^2*(d - e*x) + b^2*c^2*d*x*(8*d^2 - 55*d*e*
x + 10*e^2*x^2) + b^3*c*(-2*d^3 - 13*d^2*e*x + 16*d*e^2*x^2 + e^3*x^3)))/(c*x^2*(b + c*x)^2) - d^(3/2)*(48*c^2
*d^2 - 84*b*c*d*e + 35*b^2*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + (Sqrt[c*d - b*e]*(48*c^3*d^3 - 60*b*c^2*d^2*e
+ 11*b^2*c*d*e^2 + b^3*e^3)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/c^(3/2))/(4*b^5)

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Maple [B]  time = 0.226, size = 627, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^4/b/(c*e*x+b*e)^2*(e*x+d)^(3/2)+5/2*e^3/b^2/(c*e*x+b*e)^2*(e*x+d)^(3/2)*c*d-23/4*e^2/b^3/(c*e*x+b*e)^2*(
e*x+d)^(3/2)*c^2*d^2+3*e/b^4/(c*e*x+b*e)^2*(e*x+d)^(3/2)*c^3*d^3-1/4*e^5/(c*e*x+b*e)^2/c*(e*x+d)^(1/2)+15/4*e^
4/b/(c*e*x+b*e)^2*(e*x+d)^(1/2)*d-39/4*e^3/b^2/(c*e*x+b*e)^2*c*(e*x+d)^(1/2)*d^2+37/4*e^2/b^3/(c*e*x+b*e)^2*(e
*x+d)^(1/2)*d^3*c^2-3*e/b^4/(c*e*x+b*e)^2*c^3*(e*x+d)^(1/2)*d^4+1/4*e^4/b/c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)
^(1/2)*c/((b*e-c*d)*c)^(1/2))+5/2*e^3/b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d-71
/4*e^2/b^3*c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^2+27*e/b^4/((b*e-c*d)*c)^(1/2)*
arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^3*c^2-12/b^5*c^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*
e-c*d)*c)^(1/2))*d^4-13/4*d^2/b^3/x^2*(e*x+d)^(3/2)+3/e*d^3/b^4/x^2*(e*x+d)^(3/2)*c+11/4*d^3/b^3/x^2*(e*x+d)^(
1/2)-3/e*d^4/b^4/x^2*c*(e*x+d)^(1/2)-35/4*e^2*d^(3/2)/b^3*arctanh((e*x+d)^(1/2)/d^(1/2))+21*e*d^(5/2)/b^4*arct
anh((e*x+d)^(1/2)/d^(1/2))*c-12*d^(7/2)/b^5*arctanh((e*x+d)^(1/2)/d^(1/2))*c^2

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.15726, size = 4234, normalized size = 17.07 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(((48*c^5*d^3 - 60*b*c^4*d^2*e + 11*b^2*c^3*d*e^2 + b^3*c^2*e^3)*x^4 + 2*(48*b*c^4*d^3 - 60*b^2*c^3*d^2*e
+ 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 + (48*b^2*c^3*d^3 - 60*b^3*c^2*d^2*e + 11*b^4*c*d*e^2 + b^5*e^3)*x^2)*sqr
t((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + ((48*c^5*d^3 -
84*b*c^4*d^2*e + 35*b^2*c^3*d*e^2)*x^4 + 2*(48*b*c^4*d^3 - 84*b^2*c^3*d^2*e + 35*b^3*c^2*d*e^2)*x^3 + (48*b^2
*c^3*d^3 - 84*b^3*c^2*d^2*e + 35*b^4*c*d*e^2)*x^2)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(2
*b^4*c*d^3 - (24*b*c^4*d^3 - 36*b^2*c^3*d^2*e + 10*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 - (36*b^2*c^3*d^3 - 55*b^3*c
^2*d^2*e + 16*b^4*c*d*e^2 - b^5*e^3)*x^2 - (8*b^3*c^2*d^3 - 13*b^4*c*d^2*e)*x)*sqrt(e*x + d))/(b^5*c^3*x^4 + 2
*b^6*c^2*x^3 + b^7*c*x^2), 1/8*(2*((48*c^5*d^3 - 60*b*c^4*d^2*e + 11*b^2*c^3*d*e^2 + b^3*c^2*e^3)*x^4 + 2*(48*
b*c^4*d^3 - 60*b^2*c^3*d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 + (48*b^2*c^3*d^3 - 60*b^3*c^2*d^2*e + 11*b^4
*c*d*e^2 + b^5*e^3)*x^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + ((48
*c^5*d^3 - 84*b*c^4*d^2*e + 35*b^2*c^3*d*e^2)*x^4 + 2*(48*b*c^4*d^3 - 84*b^2*c^3*d^2*e + 35*b^3*c^2*d*e^2)*x^3
+ (48*b^2*c^3*d^3 - 84*b^3*c^2*d^2*e + 35*b^4*c*d*e^2)*x^2)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)
/x) - 2*(2*b^4*c*d^3 - (24*b*c^4*d^3 - 36*b^2*c^3*d^2*e + 10*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 - (36*b^2*c^3*d^3
- 55*b^3*c^2*d^2*e + 16*b^4*c*d*e^2 - b^5*e^3)*x^2 - (8*b^3*c^2*d^3 - 13*b^4*c*d^2*e)*x)*sqrt(e*x + d))/(b^5*c
^3*x^4 + 2*b^6*c^2*x^3 + b^7*c*x^2), 1/8*(2*((48*c^5*d^3 - 84*b*c^4*d^2*e + 35*b^2*c^3*d*e^2)*x^4 + 2*(48*b*c^
4*d^3 - 84*b^2*c^3*d^2*e + 35*b^3*c^2*d*e^2)*x^3 + (48*b^2*c^3*d^3 - 84*b^3*c^2*d^2*e + 35*b^4*c*d*e^2)*x^2)*s
qrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + ((48*c^5*d^3 - 60*b*c^4*d^2*e + 11*b^2*c^3*d*e^2 + b^3*c^2*e^3)*x^4
+ 2*(48*b*c^4*d^3 - 60*b^2*c^3*d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 + (48*b^2*c^3*d^3 - 60*b^3*c^2*d^2*e
+ 11*b^4*c*d*e^2 + b^5*e^3)*x^2)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d -
b*e)/c))/(c*x + b)) - 2*(2*b^4*c*d^3 - (24*b*c^4*d^3 - 36*b^2*c^3*d^2*e + 10*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 -
(36*b^2*c^3*d^3 - 55*b^3*c^2*d^2*e + 16*b^4*c*d*e^2 - b^5*e^3)*x^2 - (8*b^3*c^2*d^3 - 13*b^4*c*d^2*e)*x)*sqrt
(e*x + d))/(b^5*c^3*x^4 + 2*b^6*c^2*x^3 + b^7*c*x^2), 1/4*(((48*c^5*d^3 - 60*b*c^4*d^2*e + 11*b^2*c^3*d*e^2 +
b^3*c^2*e^3)*x^4 + 2*(48*b*c^4*d^3 - 60*b^2*c^3*d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 + (48*b^2*c^3*d^3 -
60*b^3*c^2*d^2*e + 11*b^4*c*d*e^2 + b^5*e^3)*x^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*
e)/c)/(c*d - b*e)) + ((48*c^5*d^3 - 84*b*c^4*d^2*e + 35*b^2*c^3*d*e^2)*x^4 + 2*(48*b*c^4*d^3 - 84*b^2*c^3*d^2*
e + 35*b^3*c^2*d*e^2)*x^3 + (48*b^2*c^3*d^3 - 84*b^3*c^2*d^2*e + 35*b^4*c*d*e^2)*x^2)*sqrt(-d)*arctan(sqrt(e*x
+ d)*sqrt(-d)/d) - (2*b^4*c*d^3 - (24*b*c^4*d^3 - 36*b^2*c^3*d^2*e + 10*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 - (36*
b^2*c^3*d^3 - 55*b^3*c^2*d^2*e + 16*b^4*c*d*e^2 - b^5*e^3)*x^2 - (8*b^3*c^2*d^3 - 13*b^4*c*d^2*e)*x)*sqrt(e*x
+ d))/(b^5*c^3*x^4 + 2*b^6*c^2*x^3 + b^7*c*x^2)]

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

[Out]

Timed out

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Giac [B]  time = 1.52494, size = 745, normalized size = 3. \begin{align*} \frac{{\left (48 \, c^{2} d^{4} - 84 \, b c d^{3} e + 35 \, b^{2} d^{2} e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{4 \, b^{5} \sqrt{-d}} - \frac{{\left (48 \, c^{4} d^{4} - 108 \, b c^{3} d^{3} e + 71 \, b^{2} c^{2} d^{2} e^{2} - 10 \, b^{3} c d e^{3} - b^{4} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{4 \, \sqrt{-c^{2} d + b c e} b^{5} c} + \frac{24 \,{\left (x e + d\right )}^{\frac{7}{2}} c^{4} d^{3} e - 72 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{4} d^{4} e + 72 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{4} d^{5} e - 24 \, \sqrt{x e + d} c^{4} d^{6} e - 36 \,{\left (x e + d\right )}^{\frac{7}{2}} b c^{3} d^{2} e^{2} + 144 \,{\left (x e + d\right )}^{\frac{5}{2}} b c^{3} d^{3} e^{2} - 180 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{3} d^{4} e^{2} + 72 \, \sqrt{x e + d} b c^{3} d^{5} e^{2} + 10 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{2} c^{2} d e^{3} - 85 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{2} c^{2} d^{2} e^{3} + 148 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} c^{2} d^{3} e^{3} - 73 \, \sqrt{x e + d} b^{2} c^{2} d^{4} e^{3} +{\left (x e + d\right )}^{\frac{7}{2}} b^{3} c e^{4} + 13 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} c d e^{4} - 42 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} c d^{2} e^{4} + 26 \, \sqrt{x e + d} b^{3} c d^{3} e^{4} -{\left (x e + d\right )}^{\frac{5}{2}} b^{4} e^{5} + 2 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d e^{5} - \sqrt{x e + d} b^{4} d^{2} e^{5}}{4 \,{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )}^{2} b^{4} c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(48*c^2*d^4 - 84*b*c*d^3*e + 35*b^2*d^2*e^2)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^5*sqrt(-d)) - 1/4*(48*c^4*d
^4 - 108*b*c^3*d^3*e + 71*b^2*c^2*d^2*e^2 - 10*b^3*c*d*e^3 - b^4*e^4)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c
*e))/(sqrt(-c^2*d + b*c*e)*b^5*c) + 1/4*(24*(x*e + d)^(7/2)*c^4*d^3*e - 72*(x*e + d)^(5/2)*c^4*d^4*e + 72*(x*e
+ d)^(3/2)*c^4*d^5*e - 24*sqrt(x*e + d)*c^4*d^6*e - 36*(x*e + d)^(7/2)*b*c^3*d^2*e^2 + 144*(x*e + d)^(5/2)*b*
c^3*d^3*e^2 - 180*(x*e + d)^(3/2)*b*c^3*d^4*e^2 + 72*sqrt(x*e + d)*b*c^3*d^5*e^2 + 10*(x*e + d)^(7/2)*b^2*c^2*
d*e^3 - 85*(x*e + d)^(5/2)*b^2*c^2*d^2*e^3 + 148*(x*e + d)^(3/2)*b^2*c^2*d^3*e^3 - 73*sqrt(x*e + d)*b^2*c^2*d^
4*e^3 + (x*e + d)^(7/2)*b^3*c*e^4 + 13*(x*e + d)^(5/2)*b^3*c*d*e^4 - 42*(x*e + d)^(3/2)*b^3*c*d^2*e^4 + 26*sqr
t(x*e + d)*b^3*c*d^3*e^4 - (x*e + d)^(5/2)*b^4*e^5 + 2*(x*e + d)^(3/2)*b^4*d*e^5 - sqrt(x*e + d)*b^4*d^2*e^5)/
(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)^2*b^4*c)