3.1744 \(\int \frac{(A+B x) (d+e x)^{3/2}}{a+b x} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 (d+e x)^{3/2} (A b-a B)}{3 b^2}+\frac{2 \sqrt{d+e x} (A b-a B) (b d-a e)}{b^3}-\frac{2 (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}+\frac{2 B (d+e x)^{5/2}}{5 b e} \]

[Out]

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

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Rubi [A]  time = 0.0764276, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \[ \frac{2 (d+e x)^{3/2} (A b-a B)}{3 b^2}+\frac{2 \sqrt{d+e x} (A b-a B) (b d-a e)}{b^3}-\frac{2 (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}+\frac{2 B (d+e x)^{5/2}}{5 b e} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(3/2))/(a + b*x),x]

[Out]

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

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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) (d+e x)^{3/2}}{a+b x} \, dx &=\frac{2 B (d+e x)^{5/2}}{5 b e}+\frac{\left (2 \left (\frac{5 A b e}{2}-\frac{5 a B e}{2}\right )\right ) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{5 b e}\\ &=\frac{2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac{2 B (d+e x)^{5/2}}{5 b e}+\frac{((A b-a B) (b d-a e)) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{b^2}\\ &=\frac{2 (A b-a B) (b d-a e) \sqrt{d+e x}}{b^3}+\frac{2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac{2 B (d+e x)^{5/2}}{5 b e}+\frac{\left ((A b-a B) (b d-a e)^2\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{b^3}\\ &=\frac{2 (A b-a B) (b d-a e) \sqrt{d+e x}}{b^3}+\frac{2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac{2 B (d+e x)^{5/2}}{5 b e}+\frac{\left (2 (A b-a B) (b d-a e)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^3 e}\\ &=\frac{2 (A b-a B) (b d-a e) \sqrt{d+e x}}{b^3}+\frac{2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac{2 B (d+e x)^{5/2}}{5 b e}-\frac{2 (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.183098, size = 109, normalized size = 0.84 \[ \frac{2 (A b-a B) \left (\sqrt{b} \sqrt{d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )}{3 b^{7/2}}+\frac{2 B (d+e x)^{5/2}}{5 b e} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(3/2))/(a + b*x),x]

[Out]

(2*B*(d + e*x)^(5/2))/(5*b*e) + (2*(A*b - a*B)*(Sqrt[b]*Sqrt[d + e*x]*(4*b*d - 3*a*e + b*e*x) - 3*(b*d - a*e)^
(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]))/(3*b^(7/2))

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Maple [B]  time = 0.009, size = 370, normalized size = 2.9 \begin{align*}{\frac{2\,B}{5\,be} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{2\,A}{3\,b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{2\,Ba}{3\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-2\,{\frac{Aae\sqrt{ex+d}}{{b}^{2}}}+2\,{\frac{Ad\sqrt{ex+d}}{b}}+2\,{\frac{{a}^{2}eB\sqrt{ex+d}}{{b}^{3}}}-2\,{\frac{Bad\sqrt{ex+d}}{{b}^{2}}}+2\,{\frac{{e}^{2}A{a}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-4\,{\frac{aAde}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+2\,{\frac{A{d}^{2}}{\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-2\,{\frac{B{a}^{3}{e}^{2}}{{b}^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+4\,{\frac{{a}^{2}eBd}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-2\,{\frac{Ba{d}^{2}}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55856, size = 795, normalized size = 6.12 \begin{align*} \left [-\frac{15 \,{\left ({\left (B a b - A b^{2}\right )} d e -{\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) - 2 \,{\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \,{\left (B a b - A b^{2}\right )} d e + 15 \,{\left (B a^{2} - A a b\right )} e^{2} +{\left (6 \, B b^{2} d e - 5 \,{\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \, b^{3} e}, \frac{2 \,{\left (15 \,{\left ({\left (B a b - A b^{2}\right )} d e -{\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) +{\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \,{\left (B a b - A b^{2}\right )} d e + 15 \,{\left (B a^{2} - A a b\right )} e^{2} +{\left (6 \, B b^{2} d e - 5 \,{\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt{e x + d}\right )}}{15 \, b^{3} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*((B*a*b - A*b^2)*d*e - (B*a^2 - A*a*b)*e^2)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e
*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(3*B*b^2*e^2*x^2 + 3*B*b^2*d^2 - 20*(B*a*b - A*b^2)*d*e + 15*(B*
a^2 - A*a*b)*e^2 + (6*B*b^2*d*e - 5*(B*a*b - A*b^2)*e^2)*x)*sqrt(e*x + d))/(b^3*e), 2/15*(15*((B*a*b - A*b^2)*
d*e - (B*a^2 - A*a*b)*e^2)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (3
*B*b^2*e^2*x^2 + 3*B*b^2*d^2 - 20*(B*a*b - A*b^2)*d*e + 15*(B*a^2 - A*a*b)*e^2 + (6*B*b^2*d*e - 5*(B*a*b - A*b
^2)*e^2)*x)*sqrt(e*x + d))/(b^3*e)]

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Sympy [A]  time = 31.0056, size = 139, normalized size = 1.07 \begin{align*} \frac{2 B \left (d + e x\right )^{\frac{5}{2}}}{5 b e} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (2 A b - 2 B a\right )}{3 b^{2}} + \frac{\sqrt{d + e x} \left (- 2 A a b e + 2 A b^{2} d + 2 B a^{2} e - 2 B a b d\right )}{b^{3}} - \frac{2 \left (- A b + B a\right ) \left (a e - b d\right )^{2} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{a e - b d}{b}}} \right )}}{b^{4} \sqrt{\frac{a e - b d}{b}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*(d + e*x)**(5/2)/(5*b*e) + (d + e*x)**(3/2)*(2*A*b - 2*B*a)/(3*b**2) + sqrt(d + e*x)*(-2*A*a*b*e + 2*A*b**
2*d + 2*B*a**2*e - 2*B*a*b*d)/b**3 - 2*(-A*b + B*a)*(a*e - b*d)**2*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(b*
*4*sqrt((a*e - b*d)/b))

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Giac [B]  time = 2.2389, size = 308, normalized size = 2.37 \begin{align*} -\frac{2 \,{\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{3}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{4} e^{4} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{3} e^{5} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{4} e^{5} - 15 \, \sqrt{x e + d} B a b^{3} d e^{5} + 15 \, \sqrt{x e + d} A b^{4} d e^{5} + 15 \, \sqrt{x e + d} B a^{2} b^{2} e^{6} - 15 \, \sqrt{x e + d} A a b^{3} e^{6}\right )} e^{\left (-5\right )}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a*b^2*d^2 - A*b^3*d^2 - 2*B*a^2*b*d*e + 2*A*a*b^2*d*e + B*a^3*e^2 - A*a^2*b*e^2)*arctan(sqrt(x*e + d)*b/
sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3) + 2/15*(3*(x*e + d)^(5/2)*B*b^4*e^4 - 5*(x*e + d)^(3/2)*B*a*b
^3*e^5 + 5*(x*e + d)^(3/2)*A*b^4*e^5 - 15*sqrt(x*e + d)*B*a*b^3*d*e^5 + 15*sqrt(x*e + d)*A*b^4*d*e^5 + 15*sqrt
(x*e + d)*B*a^2*b^2*e^6 - 15*sqrt(x*e + d)*A*a*b^3*e^6)*e^(-5)/b^5