### 3.361 $$\int \frac{(d+e x)^{7/2}}{b x+c x^2} \, dx$$

Optimal. Leaf size=157 $\frac{2 e \sqrt{d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c^3}+\frac{2 e (d+e x)^{3/2} (2 c d-b e)}{3 c^2}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 e (d+e x)^{5/2}}{5 c}$

[Out]

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

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Rubi [A]  time = 0.387282, antiderivative size = 157, 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 = {703, 824, 826, 1166, 208} $\frac{2 e \sqrt{d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c^3}+\frac{2 e (d+e x)^{3/2} (2 c d-b e)}{3 c^2}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 e (d+e x)^{5/2}}{5 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(7/2)/(b*x + c*x^2),x]

[Out]

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

Rule 703

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

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), 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] && FractionQ[m] && GtQ[m, 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}}{b x+c x^2} \, dx &=\frac{2 e (d+e x)^{5/2}}{5 c}+\frac{\int \frac{(d+e x)^{3/2} \left (c d^2+e (2 c d-b e) x\right )}{b x+c x^2} \, dx}{c}\\ &=\frac{2 e (2 c d-b e) (d+e x)^{3/2}}{3 c^2}+\frac{2 e (d+e x)^{5/2}}{5 c}+\frac{\int \frac{\sqrt{d+e x} \left (c^2 d^3+e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) x\right )}{b x+c x^2} \, dx}{c^2}\\ &=\frac{2 e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \sqrt{d+e x}}{c^3}+\frac{2 e (2 c d-b e) (d+e x)^{3/2}}{3 c^2}+\frac{2 e (d+e x)^{5/2}}{5 c}+\frac{\int \frac{c^3 d^4+e (2 c d-b e) \left (2 c^2 d^2-2 b c d e+b^2 e^2\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{c^3}\\ &=\frac{2 e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \sqrt{d+e x}}{c^3}+\frac{2 e (2 c d-b e) (d+e x)^{3/2}}{3 c^2}+\frac{2 e (d+e x)^{5/2}}{5 c}+\frac{2 \operatorname{Subst}\left (\int \frac{c^3 d^4 e-d e (2 c d-b e) \left (2 c^2 d^2-2 b c d e+b^2 e^2\right )+e (2 c d-b e) \left (2 c^2 d^2-2 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 )}{c^3}\\ &=\frac{2 e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \sqrt{d+e x}}{c^3}+\frac{2 e (2 c d-b e) (d+e x)^{3/2}}{3 c^2}+\frac{2 e (d+e x)^{5/2}}{5 c}+\frac{\left (2 c d^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}-\frac{\left (2 (c d-b e)^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^3}\\ &=\frac{2 e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \sqrt{d+e x}}{c^3}+\frac{2 e (2 c d-b e) (d+e x)^{3/2}}{3 c^2}+\frac{2 e (d+e x)^{5/2}}{5 c}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.206247, size = 138, normalized size = 0.88 $\frac{2 e \sqrt{d+e x} \left (15 b^2 e^2-5 b c e (10 d+e x)+c^2 \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )}{15 c^3}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*e*Sqrt[d + e*x]*(15*b^2*e^2 - 5*b*c*e*(10*d + e*x) + c^2*(58*d^2 + 16*d*e*x + 3*e^2*x^2)))/(15*c^3) - (2*d^
(7/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b + (2*(c*d - b*e)^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]
])/(b*c^(7/2))

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Maple [B]  time = 0.246, size = 336, normalized size = 2.1 \begin{align*}{\frac{2\,e}{5\,c} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{2\,b{e}^{2}}{3\,{c}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{4\,de}{3\,c} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{e}^{3}{b}^{2}\sqrt{ex+d}}{{c}^{3}}}-6\,{\frac{bd{e}^{2}\sqrt{ex+d}}{{c}^{2}}}+6\,{\frac{e{d}^{2}\sqrt{ex+d}}{c}}-2\,{\frac{{b}^{3}{e}^{4}}{{c}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+8\,{\frac{{e}^{3}{b}^{2}d}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-12\,{\frac{b{e}^{2}{d}^{2}}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+8\,{\frac{e{d}^{3}}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{c{d}^{4}}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{{d}^{7/2}}{b}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*e*(e*x+d)^(5/2)/c-2/3/c^2*(e*x+d)^(3/2)*b*e^2+4/3*e/c*(e*x+d)^(3/2)*d+2/c^3*b^2*e^3*(e*x+d)^(1/2)-6/c^2*b*
d*e^2*(e*x+d)^(1/2)+6*e/c*d^2*(e*x+d)^(1/2)-2/c^3*b^3*e^4/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d
)*c)^(1/2))+8/c^2*b^2*e^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d-12/c*b*e^2/((b*e-c
*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^2+8*e/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((
b*e-c*d)*c)^(1/2))*d^3-2*c/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^4-2*d^(7/2)*arc
tanh((e*x+d)^(1/2)/d^(1/2))/b

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.5173, size = 1820, normalized size = 11.59 \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),x, algorithm="fricas")

[Out]

[1/15*(15*c^3*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 15*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e
^2 - b^3*e^3)*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*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*sqrt(
e*x + d))/(b*c^3), 1/15*(15*c^3*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 30*(c^3*d^3 - 3*b*c^2*d
^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e))
+ 2*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*sqrt(
e*x + d))/(b*c^3), 1/15*(30*c^3*sqrt(-d)*d^3*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 15*(c^3*d^3 - 3*b*c^2*d^2*e +
3*b^2*c*d*e^2 - b^3*e^3)*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*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^
3)*x)*sqrt(e*x + d))/(b*c^3), 2/15*(15*c^3*sqrt(-d)*d^3*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 15*(c^3*d^3 - 3*b*c
^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b
*e)) + (3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*sqr
t(e*x + d))/(b*c^3)]

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Sympy [A]  time = 92.468, size = 162, normalized size = 1.03 \begin{align*} \frac{2 e \left (d + e x\right )^{\frac{5}{2}}}{5 c} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (- 2 b e^{2} + 4 c d e\right )}{3 c^{2}} + \frac{\sqrt{d + e x} \left (2 b^{2} e^{3} - 6 b c d e^{2} + 6 c^{2} d^{2} e\right )}{c^{3}} + \frac{2 d^{4} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b \sqrt{- d}} - \frac{2 \left (b e - c d\right )^{4} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b c^{4} \sqrt{\frac{b e - c d}{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),x)

[Out]

2*e*(d + e*x)**(5/2)/(5*c) + (d + e*x)**(3/2)*(-2*b*e**2 + 4*c*d*e)/(3*c**2) + sqrt(d + e*x)*(2*b**2*e**3 - 6*
b*c*d*e**2 + 6*c**2*d**2*e)/c**3 + 2*d**4*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) - 2*(b*e - c*d)**4*atan(sq
rt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**4*sqrt((b*e - c*d)/c))

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Giac [A]  time = 1.25215, size = 309, normalized size = 1.97 \begin{align*} \frac{2 \, d^{4} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d}} - \frac{2 \,{\left (c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 4 \, 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 )}{\sqrt{-c^{2} d + b c e} b c^{3}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{4} e + 10 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{4} d e + 45 \, \sqrt{x e + d} c^{4} d^{2} e - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{3} e^{2} - 45 \, \sqrt{x e + d} b c^{3} d e^{2} + 15 \, \sqrt{x e + d} b^{2} c^{2} e^{3}\right )}}{15 \, c^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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