3.2247 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac {(-3 a B e+2 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2} \sqrt {e}}+\frac {\sqrt {a+b x} \sqrt {d+e x} (-3 a B e+2 A b e+b B d)}{b^2 (b d-a e)}-\frac {2 (d+e x)^{3/2} (A b-a B)}{b \sqrt {a+b x} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {78, 50, 63, 217, 206} \[ \frac {\sqrt {a+b x} \sqrt {d+e x} (-3 a B e+2 A b e+b B d)}{b^2 (b d-a e)}+\frac {(-3 a B e+2 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2} \sqrt {e}}-\frac {2 (d+e x)^{3/2} (A b-a B)}{b \sqrt {a+b x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]  time = 0.49, size = 125, normalized size = 0.86 \[ \frac {\frac {b (d+e x) (3 a B-2 A b+b B x)}{\sqrt {a+b x}}+\frac {\sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}} (-3 a B e+2 A b e+b B d) \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{\sqrt {e}}}{b^3 \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.72, size = 366, normalized size = 2.52 \[ \left [\frac {{\left (B a b d - {\left (3 \, B a^{2} - 2 \, A a b\right )} e + {\left (B b^{2} d - {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (B b^{2} e x + {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {e x + d}}{4 \, {\left (b^{4} e x + a b^{3} e\right )}}, -\frac {{\left (B a b d - {\left (3 \, B a^{2} - 2 \, A a b\right )} e + {\left (B b^{2} d - {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \, {\left (B b^{2} e x + {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{4} e x + a b^{3} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.45, size = 227, normalized size = 1.57 \[ \frac {\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} B {\left | b \right |}}{b^{4}} - \frac {{\left (B b^{\frac {3}{2}} d {\left | b \right |} e^{\frac {1}{2}} - 3 \, B a \sqrt {b} {\left | b \right |} e^{\frac {3}{2}} + 2 \, A b^{\frac {3}{2}} {\left | b \right |} e^{\frac {3}{2}}\right )} e^{\left (-1\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}{2 \, b^{4}} + \frac {4 \, {\left (B a b^{\frac {3}{2}} d {\left | b \right |} e^{\frac {1}{2}} - A b^{\frac {5}{2}} d {\left | b \right |} e^{\frac {1}{2}} - B a^{2} \sqrt {b} {\left | b \right |} e^{\frac {3}{2}} + A a b^{\frac {3}{2}} {\left | b \right |} e^{\frac {3}{2}}\right )}}{{\left (b^{2} d - a b e - {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 386, normalized size = 2.66 \[ \frac {\sqrt {e x +d}\, \left (2 A \,b^{2} e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-3 B a b e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+B \,b^{2} d x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+2 A a b e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-3 B \,a^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+B a b d \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b x -4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a \right )}{2 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b x +a}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e*x+d)^(1/2)*(2*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*b^2*e-3*B
*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*a*b*e+B*ln(1/2*(2*b*e*x+a*e+b*d
+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*b^2*d+2*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(
1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a*b*e-3*B*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^
(1/2))*a^2*e+B*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a*b*d+2*B*x*b*((b*x
+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-4*A*b*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*B*a*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(
1/2))/(b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/b^2/(b*x+a)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,x\right )\,\sqrt {d+e\,x}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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