3.1762 \(\int \frac{A+B x}{(a+b x)^3 \sqrt{d+e x}} \, dx\)

Optimal. Leaf size=157 \[ \frac{e (-a B e-3 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{5/2}}-\frac{\sqrt{d+e x} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}-\frac{\sqrt{d+e x} (-a B e-3 A b e+4 b B d)}{4 b (a+b x) (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.124942, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {78, 51, 63, 208} \[ \frac{e (-a B e-3 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{5/2}}-\frac{\sqrt{d+e x} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}-\frac{\sqrt{d+e x} (-a B e-3 A b e+4 b B d)}{4 b (a+b x) (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)^3*Sqrt[d + e*x]),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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}{(a+b x)^3 \sqrt{d+e x}} \, dx &=-\frac{(A b-a B) \sqrt{d+e x}}{2 b (b d-a e) (a+b x)^2}+\frac{(4 b B d-3 A b e-a B e) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{4 b (b d-a e)}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{2 b (b d-a e) (a+b x)^2}-\frac{(4 b B d-3 A b e-a B e) \sqrt{d+e x}}{4 b (b d-a e)^2 (a+b x)}-\frac{(e (4 b B d-3 A b e-a B e)) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{2 b (b d-a e) (a+b x)^2}-\frac{(4 b B d-3 A b e-a B e) \sqrt{d+e x}}{4 b (b d-a e)^2 (a+b x)}-\frac{(4 b B d-3 A b e-a B e) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 b (b d-a e)^2}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{2 b (b d-a e) (a+b x)^2}-\frac{(4 b B d-3 A b e-a B e) \sqrt{d+e x}}{4 b (b d-a e)^2 (a+b x)}+\frac{e (4 b B d-3 A b e-a B e) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.289756, size = 149, normalized size = 0.95 \[ \frac{\sqrt{d+e x} \left (\frac{e (-a B e-3 A b e+4 b B d) \left (\frac{a e-b d}{e (a+b x)}+\frac{\sqrt{a e-b d} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{\sqrt{b} \sqrt{d+e x}}\right )}{2 (b d-a e)^2}+\frac{a B-A b}{(a+b x)^2}\right )}{2 b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)^3*Sqrt[d + e*x]),x]

[Out]

(Sqrt[d + e*x]*((-(A*b) + a*B)/(a + b*x)^2 + (e*(4*b*B*d - 3*A*b*e - a*B*e)*((-(b*d) + a*e)/(e*(a + b*x)) + (S
qrt[-(b*d) + a*e]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(Sqrt[b]*Sqrt[d + e*x])))/(2*(b*d - a*e)
^2)))/(2*b*(b*d - a*e))

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Maple [B]  time = 0.016, size = 436, normalized size = 2.8 \begin{align*}{\frac{3\,Ab{e}^{2}}{4\, \left ( bxe+ae \right ) ^{2} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{Ba{e}^{2}}{4\, \left ( bxe+ae \right ) ^{2} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{bBde}{ \left ( bxe+ae \right ) ^{2} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{e}^{2}}{4\, \left ( bxe+ae \right ) ^{2} \left ( ae-bd \right ) }\sqrt{ex+d}}-{\frac{Ba{e}^{2}}{4\, \left ( bxe+ae \right ) ^{2} \left ( ae-bd \right ) b}\sqrt{ex+d}}-{\frac{eBd}{ \left ( bxe+ae \right ) ^{2} \left ( ae-bd \right ) }\sqrt{ex+d}}+{\frac{3\,A{e}^{2}}{4\,{a}^{2}{e}^{2}-8\,abde+4\,{b}^{2}{d}^{2}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{Ba{e}^{2}}{ \left ( 4\,{a}^{2}{e}^{2}-8\,abde+4\,{b}^{2}{d}^{2} \right ) b}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}-{\frac{eBd}{{a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^3/(e*x+d)^(1/2),x)

[Out]

3/4/(b*e*x+a*e)^2/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)*A*b*e^2+1/4/(b*e*x+a*e)^2/(a^2*e^2-2*a*b*d*e+b^2*d
^2)*(e*x+d)^(3/2)*B*a*e^2-e/(b*e*x+a*e)^2/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)*B*b*d+5/4/(b*e*x+a*e)^2/(a
*e-b*d)*(e*x+d)^(1/2)*A*e^2-1/4/(b*e*x+a*e)^2/(a*e-b*d)/b*(e*x+d)^(1/2)*B*a*e^2-e/(b*e*x+a*e)^2/(a*e-b*d)*(e*x
+d)^(1/2)*B*d+3/4/(a^2*e^2-2*a*b*d*e+b^2*d^2)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*
A*e^2+1/4/(a^2*e^2-2*a*b*d*e+b^2*d^2)/b/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*B*a*e^
2-e/(a^2*e^2-2*a*b*d*e+b^2*d^2)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*B*d

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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)/(b*x+a)^3/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.4934, size = 1646, normalized size = 10.48 \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)/(b*x+a)^3/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((4*B*a^2*b*d*e - (B*a^3 + 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (B*a*b^2 + 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d
*e - (B*a^2*b + 3*A*a*b^2)*e^2)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e
*x + d))/(b*x + a)) + 2*(2*(B*a*b^3 + A*b^4)*d^2 - (B*a^2*b^2 + 7*A*a*b^3)*d*e - (B*a^3*b - 5*A*a^2*b^2)*e^2 +
(4*B*b^4*d^2 - (5*B*a*b^3 + 3*A*b^4)*d*e + (B*a^2*b^2 + 3*A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^3 - 3*a^
3*b^4*d^2*e + 3*a^4*b^3*d*e^2 - a^5*b^2*e^3 + (b^7*d^3 - 3*a*b^6*d^2*e + 3*a^2*b^5*d*e^2 - a^3*b^4*e^3)*x^2 +
2*(a*b^6*d^3 - 3*a^2*b^5*d^2*e + 3*a^3*b^4*d*e^2 - a^4*b^3*e^3)*x), -1/4*((4*B*a^2*b*d*e - (B*a^3 + 3*A*a^2*b)
*e^2 + (4*B*b^3*d*e - (B*a*b^2 + 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (B*a^2*b + 3*A*a*b^2)*e^2)*x)*sqrt(-b^
2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (2*(B*a*b^3 + A*b^4)*d^2 - (B*a^2*b^2
+ 7*A*a*b^3)*d*e - (B*a^3*b - 5*A*a^2*b^2)*e^2 + (4*B*b^4*d^2 - (5*B*a*b^3 + 3*A*b^4)*d*e + (B*a^2*b^2 + 3*A*a
*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^3 - 3*a^3*b^4*d^2*e + 3*a^4*b^3*d*e^2 - a^5*b^2*e^3 + (b^7*d^3 - 3*a*b
^6*d^2*e + 3*a^2*b^5*d*e^2 - a^3*b^4*e^3)*x^2 + 2*(a*b^6*d^3 - 3*a^2*b^5*d^2*e + 3*a^3*b^4*d*e^2 - a^4*b^3*e^3
)*x)]

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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*x+A)/(b*x+a)**3/(e*x+d)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.17817, size = 359, normalized size = 2.29 \begin{align*} -\frac{{\left (4 \, B b d e - B a e^{2} - 3 \, A b e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \,{\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} \sqrt{-b^{2} d + a b e}} - \frac{4 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{2} d e - 4 \, \sqrt{x e + d} B b^{2} d^{2} e -{\left (x e + d\right )}^{\frac{3}{2}} B a b e^{2} - 3 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{2} e^{2} + 3 \, \sqrt{x e + d} B a b d e^{2} + 5 \, \sqrt{x e + d} A b^{2} d e^{2} + \sqrt{x e + d} B a^{2} e^{3} - 5 \, \sqrt{x e + d} A a b e^{3}}{4 \,{\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*B*b*d*e - B*a*e^2 - 3*A*b*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^2 - 2*a*b^2*d*e +
a^2*b*e^2)*sqrt(-b^2*d + a*b*e)) - 1/4*(4*(x*e + d)^(3/2)*B*b^2*d*e - 4*sqrt(x*e + d)*B*b^2*d^2*e - (x*e + d)^
(3/2)*B*a*b*e^2 - 3*(x*e + d)^(3/2)*A*b^2*e^2 + 3*sqrt(x*e + d)*B*a*b*d*e^2 + 5*sqrt(x*e + d)*A*b^2*d*e^2 + sq
rt(x*e + d)*B*a^2*e^3 - 5*sqrt(x*e + d)*A*a*b*e^3)/((b^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2)*((x*e + d)*b - b*d + a
*e)^2)