3.1235 \(\int \frac{A+B x}{(d+e x)^{5/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=164 \[ -\frac{2 c^{3/2} (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{5/2}}+\frac{2 \left (B c d^2-A e (2 c d-b e)\right )}{d^2 \sqrt{d+e x} (c d-b e)^2}+\frac{2 (B d-A e)}{3 d (d+e x)^{3/2} (c d-b e)}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.324719, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {828, 826, 1166, 208} \[ -\frac{2 c^{3/2} (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{5/2}}+\frac{2 \left (B c d^2-A e (2 c d-b e)\right )}{d^2 \sqrt{d+e x} (c d-b e)^2}+\frac{2 (B d-A e)}{3 d (d+e x)^{3/2} (c d-b e)}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^(5/2)*(b*x + c*x^2)),x]

[Out]

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

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

Mathematica [C]  time = 0.032829, size = 91, normalized size = 0.55 \[ \frac{2 \left (d (b B-A c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c (d+e x)}{c d-b e}\right )+A (c d-b e) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{e x}{d}+1\right )\right )}{3 b d (d+e x)^{3/2} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)^(5/2)*(b*x + c*x^2)),x]

[Out]

(2*((b*B - A*c)*d*Hypergeometric2F1[-3/2, 1, -1/2, (c*(d + e*x))/(c*d - b*e)] + A*(c*d - b*e)*Hypergeometric2F
1[-3/2, 1, -1/2, 1 + (e*x)/d]))/(3*b*d*(c*d - b*e)*(d + e*x)^(3/2))

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Maple [A]  time = 0.019, size = 243, normalized size = 1.5 \begin{align*}{\frac{2\,Ae}{3\,d \left ( be-cd \right ) } \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,B}{3\,be-3\,cd} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{Ab{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}-4\,{\frac{Ace}{d \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}+2\,{\frac{Bc}{ \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}-2\,{\frac{A}{b{d}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-2\,{\frac{{c}^{3}A}{ \left ( be-cd \right ) ^{2}b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+2\,{\frac{B{c}^{2}}{ \left ( be-cd \right ) ^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x),x)

[Out]

2/3/d/(b*e-c*d)/(e*x+d)^(3/2)*A*e-2/3/(b*e-c*d)/(e*x+d)^(3/2)*B+2/d^2/(b*e-c*d)^2/(e*x+d)^(1/2)*A*b*e^2-4/d/(b
*e-c*d)^2/(e*x+d)^(1/2)*A*c*e+2/(b*e-c*d)^2/(e*x+d)^(1/2)*B*c-2*A*arctanh((e*x+d)^(1/2)/d^(1/2))/b/d^(5/2)-2/(
b*e-c*d)^2*c^3/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A+2/(b*e-c*d)^2*c^2/((b*e-c*d
)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(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((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 10.4109, size = 3584, normalized size = 21.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[-1/3*(3*((B*b*c - A*c^2)*d^3*e^2*x^2 + 2*(B*b*c - A*c^2)*d^4*e*x + (B*b*c - A*c^2)*d^5)*sqrt(c/(c*d - b*e))*l
og((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)) - 3*(A*c^2*d^4 - 2*A*b*c
*d^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2*e^2 - 2*A*b*c*d*e^3 + A*b^2*e^4)*x^2 + 2*(A*c^2*d^3*e - 2*A*b*c*d^2*e^2 +
A*b^2*d*e^3)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(4*B*b*c*d^4 + 4*A*b^2*d^2*e^2 - (B*b
^2 + 7*A*b*c)*d^3*e + 3*(B*b*c*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c^2*d^7 - 2*b^2*c*d
^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e - 2*b^2*c*d^5*e^2 +
b^3*d^4*e^3)*x), -1/3*(6*((B*b*c - A*c^2)*d^3*e^2*x^2 + 2*(B*b*c - A*c^2)*d^4*e*x + (B*b*c - A*c^2)*d^5)*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)) - 3*(A*c^2*d^4 - 2*A*b*c
*d^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2*e^2 - 2*A*b*c*d*e^3 + A*b^2*e^4)*x^2 + 2*(A*c^2*d^3*e - 2*A*b*c*d^2*e^2 +
A*b^2*d*e^3)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(4*B*b*c*d^4 + 4*A*b^2*d^2*e^2 - (B*b
^2 + 7*A*b*c)*d^3*e + 3*(B*b*c*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c^2*d^7 - 2*b^2*c*d
^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e - 2*b^2*c*d^5*e^2 +
b^3*d^4*e^3)*x), 1/3*(6*(A*c^2*d^4 - 2*A*b*c*d^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2*e^2 - 2*A*b*c*d*e^3 + A*b^2*e^
4)*x^2 + 2*(A*c^2*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 3*((B*
b*c - A*c^2)*d^3*e^2*x^2 + 2*(B*b*c - A*c^2)*d^4*e*x + (B*b*c - A*c^2)*d^5)*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*(4*B*b*c*d^4 + 4*A*b^2*d^2*e^2 -
(B*b^2 + 7*A*b*c)*d^3*e + 3*(B*b*c*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c^2*d^7 - 2*b^2
*c*d^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e - 2*b^2*c*d^5*e^
2 + b^3*d^4*e^3)*x), -2/3*(3*((B*b*c - A*c^2)*d^3*e^2*x^2 + 2*(B*b*c - A*c^2)*d^4*e*x + (B*b*c - A*c^2)*d^5)*s
qrt(-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)) - 3*(A*c^2*d^4 - 2*A
*b*c*d^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2*e^2 - 2*A*b*c*d*e^3 + A*b^2*e^4)*x^2 + 2*(A*c^2*d^3*e - 2*A*b*c*d^2*e^
2 + A*b^2*d*e^3)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (4*B*b*c*d^4 + 4*A*b^2*d^2*e^2 - (B*b^2 + 7*A*
b*c)*d^3*e + 3*(B*b*c*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c^2*d^7 - 2*b^2*c*d^6*e + b^
3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e - 2*b^2*c*d^5*e^2 + b^3*d^4*e
^3)*x)]

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Sympy [A]  time = 51.427, size = 160, normalized size = 0.98 \begin{align*} \frac{2 A \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b d^{2} \sqrt{- d}} - \frac{2 \left (- A e + B d\right )}{3 d \left (d + e x\right )^{\frac{3}{2}} \left (b e - c d\right )} + \frac{2 \left (A b e^{2} - 2 A c d e + B c d^{2}\right )}{d^{2} \sqrt{d + e x} \left (b e - c d\right )^{2}} + \frac{2 c \left (- A c + B b\right ) \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 )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**(5/2)/(c*x**2+b*x),x)

[Out]

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

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Giac [A]  time = 1.27681, size = 293, normalized size = 1.79 \begin{align*} \frac{2 \,{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} \sqrt{-c^{2} d + b c e}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )} B c d^{2} + B c d^{3} - 6 \,{\left (x e + d\right )} A c d e - B b d^{2} e - A c d^{2} e + 3 \,{\left (x e + d\right )} A b e^{2} + A b d e^{2}\right )}}{3 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}{\left (x e + d\right )}^{\frac{3}{2}}} + \frac{2 \, A \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

2*(B*b*c^2 - A*c^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*sqrt(-c^
2*d + b*c*e)) + 2/3*(3*(x*e + d)*B*c*d^2 + B*c*d^3 - 6*(x*e + d)*A*c*d*e - B*b*d^2*e - A*c*d^2*e + 3*(x*e + d)
*A*b*e^2 + A*b*d*e^2)/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*(x*e + d)^(3/2)) + 2*A*arctan(sqrt(x*e + d)/sqrt(
-d))/(b*sqrt(-d)*d^2)