3.368 \(\int \frac{1}{(d+e x)^{7/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=187 \[ -\frac{2 e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{d^3 \sqrt{d+e x} (c d-b e)^3}+\frac{2 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{7/2}}-\frac{2 e (2 c d-b e)}{3 d^2 (d+e x)^{3/2} (c d-b e)^2}-\frac{2 e}{5 d (d+e x)^{5/2} (c d-b e)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{7/2}} \]

[Out]

(-2*e)/(5*d*(c*d - b*e)*(d + e*x)^(5/2)) - (2*e*(2*c*d - b*e))/(3*d^2*(c*d - b*e)^2*(d + e*x)^(3/2)) - (2*e*(3
*c^2*d^2 - 3*b*c*d*e + b^2*e^2))/(d^3*(c*d - b*e)^3*Sqrt[d + e*x]) - (2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*d^(
7/2)) + (2*c^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*(c*d - b*e)^(7/2))

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Rubi [A]  time = 0.376799, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {709, 828, 826, 1166, 208} \[ -\frac{2 e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{d^3 \sqrt{d+e x} (c d-b e)^3}+\frac{2 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{7/2}}-\frac{2 e (2 c d-b e)}{3 d^2 (d+e x)^{3/2} (c d-b e)^2}-\frac{2 e}{5 d (d+e x)^{5/2} (c d-b e)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(7/2)*(b*x + c*x^2)),x]

[Out]

(-2*e)/(5*d*(c*d - b*e)*(d + e*x)^(5/2)) - (2*e*(2*c*d - b*e))/(3*d^2*(c*d - b*e)^2*(d + e*x)^(3/2)) - (2*e*(3
*c^2*d^2 - 3*b*c*d*e + b^2*e^2))/(d^3*(c*d - b*e)^3*Sqrt[d + e*x]) - (2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*d^(
7/2)) + (2*c^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*(c*d - b*e)^(7/2))

Rule 709

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d - b*e - c
*e*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, 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[m, -1]

Rule 828

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

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

Mathematica [C]  time = 0.0303762, size = 83, normalized size = 0.44 \[ -\frac{2 \left (c d \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{c (d+e x)}{c d-b e}\right )+(b e-c d) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{e x}{d}+1\right )\right )}{5 b d (d+e x)^{5/2} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(7/2)*(b*x + c*x^2)),x]

[Out]

(-2*(c*d*Hypergeometric2F1[-5/2, 1, -3/2, (c*(d + e*x))/(c*d - b*e)] + (-(c*d) + b*e)*Hypergeometric2F1[-5/2,
1, -3/2, 1 + (e*x)/d]))/(5*b*d*(c*d - b*e)*(d + e*x)^(5/2))

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Maple [A]  time = 0.235, size = 228, normalized size = 1.2 \begin{align*}{\frac{2\,b{e}^{2}}{3\,{d}^{2} \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,ce}{3\,d \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{{e}^{3}{b}^{2}}{{d}^{3} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}-6\,{\frac{bc{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}+6\,{\frac{{c}^{2}e}{d \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}+{\frac{2\,e}{5\,d \left ( be-cd \right ) } \left ( ex+d \right ) ^{-{\frac{5}{2}}}}+2\,{\frac{{c}^{4}}{ \left ( be-cd \right ) ^{3}b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{1}{b{d}^{7/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(7/2)/(c*x^2+b*x),x)

[Out]

2/3/d^2/(b*e-c*d)^2/(e*x+d)^(3/2)*b*e^2-4/3*e/d/(b*e-c*d)^2/(e*x+d)^(3/2)*c+2/d^3/(b*e-c*d)^3/(e*x+d)^(1/2)*b^
2*e^3-6/d^2/(b*e-c*d)^3/(e*x+d)^(1/2)*b*c*e^2+6*e/d/(b*e-c*d)^3/(e*x+d)^(1/2)*c^2+2/5*e/d/(b*e-c*d)/(e*x+d)^(5
/2)+2/(b*e-c*d)^3*c^4/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))-2*arctanh((e*x+d)^(1/2
)/d^(1/2))/b/d^(7/2)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.15514, size = 5191, normalized size = 27.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[-1/15*(15*(c^3*d^4*e^3*x^3 + 3*c^3*d^5*e^2*x^2 + 3*c^3*d^6*e*x + c^3*d^7)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*
c*d - b*e - 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) - 15*(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*
c*d^4*e^2 - b^3*d^3*e^3 + (c^3*d^3*e^3 - 3*b*c^2*d^2*e^4 + 3*b^2*c*d*e^5 - b^3*e^6)*x^3 + 3*(c^3*d^4*e^2 - 3*b
*c^2*d^3*e^3 + 3*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2 + 3*(c^3*d^5*e - 3*b*c^2*d^4*e^2 + 3*b^2*c*d^3*e^3 - b^3*d^2*e
^4)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(58*b*c^2*d^5*e - 66*b^2*c*d^4*e^2 + 23*b^3*d^
3*e^3 + 15*(3*b*c^2*d^3*e^3 - 3*b^2*c*d^2*e^4 + b^3*d*e^5)*x^2 + 5*(20*b*c^2*d^4*e^2 - 21*b^2*c*d^3*e^3 + 7*b^
3*d^2*e^4)*x)*sqrt(e*x + d))/(b*c^3*d^10 - 3*b^2*c^2*d^9*e + 3*b^3*c*d^8*e^2 - b^4*d^7*e^3 + (b*c^3*d^7*e^3 -
3*b^2*c^2*d^6*e^4 + 3*b^3*c*d^5*e^5 - b^4*d^4*e^6)*x^3 + 3*(b*c^3*d^8*e^2 - 3*b^2*c^2*d^7*e^3 + 3*b^3*c*d^6*e^
4 - b^4*d^5*e^5)*x^2 + 3*(b*c^3*d^9*e - 3*b^2*c^2*d^8*e^2 + 3*b^3*c*d^7*e^3 - b^4*d^6*e^4)*x), 1/15*(30*(c^3*d
^4*e^3*x^3 + 3*c^3*d^5*e^2*x^2 + 3*c^3*d^6*e*x + c^3*d^7)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x +
d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + 15*(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3 + (c^3*d^
3*e^3 - 3*b*c^2*d^2*e^4 + 3*b^2*c*d*e^5 - b^3*e^6)*x^3 + 3*(c^3*d^4*e^2 - 3*b*c^2*d^3*e^3 + 3*b^2*c*d^2*e^4 -
b^3*d*e^5)*x^2 + 3*(c^3*d^5*e - 3*b*c^2*d^4*e^2 + 3*b^2*c*d^3*e^3 - b^3*d^2*e^4)*x)*sqrt(d)*log((e*x - 2*sqrt(
e*x + d)*sqrt(d) + 2*d)/x) - 2*(58*b*c^2*d^5*e - 66*b^2*c*d^4*e^2 + 23*b^3*d^3*e^3 + 15*(3*b*c^2*d^3*e^3 - 3*b
^2*c*d^2*e^4 + b^3*d*e^5)*x^2 + 5*(20*b*c^2*d^4*e^2 - 21*b^2*c*d^3*e^3 + 7*b^3*d^2*e^4)*x)*sqrt(e*x + d))/(b*c
^3*d^10 - 3*b^2*c^2*d^9*e + 3*b^3*c*d^8*e^2 - b^4*d^7*e^3 + (b*c^3*d^7*e^3 - 3*b^2*c^2*d^6*e^4 + 3*b^3*c*d^5*e
^5 - b^4*d^4*e^6)*x^3 + 3*(b*c^3*d^8*e^2 - 3*b^2*c^2*d^7*e^3 + 3*b^3*c*d^6*e^4 - b^4*d^5*e^5)*x^2 + 3*(b*c^3*d
^9*e - 3*b^2*c^2*d^8*e^2 + 3*b^3*c*d^7*e^3 - b^4*d^6*e^4)*x), 1/15*(30*(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*
e^2 - b^3*d^3*e^3 + (c^3*d^3*e^3 - 3*b*c^2*d^2*e^4 + 3*b^2*c*d*e^5 - b^3*e^6)*x^3 + 3*(c^3*d^4*e^2 - 3*b*c^2*d
^3*e^3 + 3*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2 + 3*(c^3*d^5*e - 3*b*c^2*d^4*e^2 + 3*b^2*c*d^3*e^3 - b^3*d^2*e^4)*x)
*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 15*(c^3*d^4*e^3*x^3 + 3*c^3*d^5*e^2*x^2 + 3*c^3*d^6*e*x + c^3*d^7
)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e - 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) -
2*(58*b*c^2*d^5*e - 66*b^2*c*d^4*e^2 + 23*b^3*d^3*e^3 + 15*(3*b*c^2*d^3*e^3 - 3*b^2*c*d^2*e^4 + b^3*d*e^5)*x^
2 + 5*(20*b*c^2*d^4*e^2 - 21*b^2*c*d^3*e^3 + 7*b^3*d^2*e^4)*x)*sqrt(e*x + d))/(b*c^3*d^10 - 3*b^2*c^2*d^9*e +
3*b^3*c*d^8*e^2 - b^4*d^7*e^3 + (b*c^3*d^7*e^3 - 3*b^2*c^2*d^6*e^4 + 3*b^3*c*d^5*e^5 - b^4*d^4*e^6)*x^3 + 3*(b
*c^3*d^8*e^2 - 3*b^2*c^2*d^7*e^3 + 3*b^3*c*d^6*e^4 - b^4*d^5*e^5)*x^2 + 3*(b*c^3*d^9*e - 3*b^2*c^2*d^8*e^2 + 3
*b^3*c*d^7*e^3 - b^4*d^6*e^4)*x), 2/15*(15*(c^3*d^4*e^3*x^3 + 3*c^3*d^5*e^2*x^2 + 3*c^3*d^6*e*x + c^3*d^7)*sqr
t(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + 15*(c^3*d^6 - 3*b*c^
2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3 + (c^3*d^3*e^3 - 3*b*c^2*d^2*e^4 + 3*b^2*c*d*e^5 - b^3*e^6)*x^3 + 3*(c
^3*d^4*e^2 - 3*b*c^2*d^3*e^3 + 3*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2 + 3*(c^3*d^5*e - 3*b*c^2*d^4*e^2 + 3*b^2*c*d^3
*e^3 - b^3*d^2*e^4)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (58*b*c^2*d^5*e - 66*b^2*c*d^4*e^2 + 23*b^3
*d^3*e^3 + 15*(3*b*c^2*d^3*e^3 - 3*b^2*c*d^2*e^4 + b^3*d*e^5)*x^2 + 5*(20*b*c^2*d^4*e^2 - 21*b^2*c*d^3*e^3 + 7
*b^3*d^2*e^4)*x)*sqrt(e*x + d))/(b*c^3*d^10 - 3*b^2*c^2*d^9*e + 3*b^3*c*d^8*e^2 - b^4*d^7*e^3 + (b*c^3*d^7*e^3
- 3*b^2*c^2*d^6*e^4 + 3*b^3*c*d^5*e^5 - b^4*d^4*e^6)*x^3 + 3*(b*c^3*d^8*e^2 - 3*b^2*c^2*d^7*e^3 + 3*b^3*c*d^6
*e^4 - b^4*d^5*e^5)*x^2 + 3*(b*c^3*d^9*e - 3*b^2*c^2*d^8*e^2 + 3*b^3*c*d^7*e^3 - b^4*d^6*e^4)*x)]

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Sympy [A]  time = 48.7899, size = 182, normalized size = 0.97 \begin{align*} \frac{2 e}{5 d \left (d + e x\right )^{\frac{5}{2}} \left (b e - c d\right )} + \frac{2 e \left (b e - 2 c d\right )}{3 d^{2} \left (d + e x\right )^{\frac{3}{2}} \left (b e - c d\right )^{2}} + \frac{2 e \left (b^{2} e^{2} - 3 b c d e + 3 c^{2} d^{2}\right )}{d^{3} \sqrt{d + e x} \left (b e - c d\right )^{3}} + \frac{2 c^{3} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b \sqrt{\frac{b e - c d}{c}} \left (b e - c d\right )^{3}} + \frac{2 \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b d^{3} \sqrt{- d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(7/2)/(c*x**2+b*x),x)

[Out]

2*e/(5*d*(d + e*x)**(5/2)*(b*e - c*d)) + 2*e*(b*e - 2*c*d)/(3*d**2*(d + e*x)**(3/2)*(b*e - c*d)**2) + 2*e*(b**
2*e**2 - 3*b*c*d*e + 3*c**2*d**2)/(d**3*sqrt(d + e*x)*(b*e - c*d)**3) + 2*c**3*atan(sqrt(d + e*x)/sqrt((b*e -
c*d)/c))/(b*sqrt((b*e - c*d)/c)*(b*e - c*d)**3) + 2*atan(sqrt(d + e*x)/sqrt(-d))/(b*d**3*sqrt(-d))

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Giac [A]  time = 1.30376, size = 389, normalized size = 2.08 \begin{align*} -\frac{2 \, c^{4} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} \sqrt{-c^{2} d + b c e}} - \frac{2 \,{\left (45 \,{\left (x e + d\right )}^{2} c^{2} d^{2} e + 10 \,{\left (x e + d\right )} c^{2} d^{3} e + 3 \, c^{2} d^{4} e - 45 \,{\left (x e + d\right )}^{2} b c d e^{2} - 15 \,{\left (x e + d\right )} b c d^{2} e^{2} - 6 \, b c d^{3} e^{2} + 15 \,{\left (x e + d\right )}^{2} b^{2} e^{3} + 5 \,{\left (x e + d\right )} b^{2} d e^{3} + 3 \, b^{2} d^{2} e^{3}\right )}}{15 \,{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} + \frac{2 \, \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

-2*c^4*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b*c^3*d^3 - 3*b^2*c^2*d^2*e + 3*b^3*c*d*e^2 - b^4*e^3)*s
qrt(-c^2*d + b*c*e)) - 2/15*(45*(x*e + d)^2*c^2*d^2*e + 10*(x*e + d)*c^2*d^3*e + 3*c^2*d^4*e - 45*(x*e + d)^2*
b*c*d*e^2 - 15*(x*e + d)*b*c*d^2*e^2 - 6*b*c*d^3*e^2 + 15*(x*e + d)^2*b^2*e^3 + 5*(x*e + d)*b^2*d*e^3 + 3*b^2*
d^2*e^3)/((c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*(x*e + d)^(5/2)) + 2*arctan(sqrt(x*e + d)/
sqrt(-d))/(b*sqrt(-d)*d^3)