3.1860 \(\int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=230 \[ \frac {2 (a+b x) (d+e x)^{3/2} (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) \sqrt {d+e x} (A b-a B) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

2/3*(A*b-B*a)*(b*x+a)*(e*x+d)^(3/2)/b^2/((b*x+a)^2)^(1/2)+2/5*B*(b*x+a)*(e*x+d)^(5/2)/b/e/((b*x+a)^2)^(1/2)-2*
(A*b-B*a)*(-a*e+b*d)^(3/2)*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/((b*x+a)^2)^(1/2)+2
*(A*b-B*a)*(-a*e+b*d)*(b*x+a)*(e*x+d)^(1/2)/b^3/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 80, 50, 63, 208} \[ \frac {2 (a+b x) (d+e x)^{3/2} (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) \sqrt {d+e x} (A b-a B) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(A+B x) (d+e x)^{3/2}}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{a b+b^2 x} \, dx}{5 b^2 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right ) \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{5 b^4 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right )^2 \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{5 b^6 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 \left (b^2 d-a b e\right )^2 \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{5 b^6 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (b d-a e)^{3/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 127, normalized size = 0.55 \[ \frac {2 (a+b x) \left (\frac {5 e (A b-a B) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{3 b^{5/2}}+B (d+e x)^{5/2}\right )}{5 b e \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 373, normalized size = 1.62 \[ \left [-\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3} e}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),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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giac [A]  time = 0.22, size = 306, normalized size = 1.33 \[ -\frac {2 \, {\left (B a b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, B a^{2} b d e \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + B a^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{3} e^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{4} e^{5} \mathrm {sgn}\left (b x + a\right ) - 15 \, \sqrt {x e + d} B a b^{3} d e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} A b^{4} d e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} B a^{2} b^{2} e^{6} \mathrm {sgn}\left (b x + a\right ) - 15 \, \sqrt {x e + d} A a b^{3} e^{6} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{15 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 414, normalized size = 1.80 \[ \frac {2 \left (b x +a \right ) \left (15 A \,a^{2} b \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-30 A a \,b^{2} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 A \,b^{3} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-15 B \,a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+30 B \,a^{2} b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-15 B a \,b^{2} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A a b \,e^{2}+15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A \,b^{2} d e +15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,a^{2} e^{2}-15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B a b d e +5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} A \,b^{2} e -5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} B a b e +3 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} B \,b^{2}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(b*x+a)*(3*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^2+5*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^2*e+15*A*arcta
n((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*b*e^3-30*A*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^2*d*e^2+
15*A*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*b^3*d^2*e-5*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b*e-15*B*ar
ctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^3*e^3+30*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*b*d*e^2
-15*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^2*d^2*e-15*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*e^2+1
5*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d*e+15*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-15*B*((a*e-b*d)*b
)^(1/2)*(e*x+d)^(1/2)*a*b*d*e)/((b*x+a)^2)^(1/2)/e/b^3/((a*e-b*d)*b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{\sqrt {{\left (b x + a\right )}^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (d + e x\right )^{\frac {3}{2}}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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