3.23.47 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx\) [2247]

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

[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 = 0.07, 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 = {79, 52, 65, 223, 212} \begin {gather*} \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)} \end {gather*}

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 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

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

Rule 223

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) \text {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) \text {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 = 0.20, size = 92, normalized size = 0.63 \begin {gather*} \frac {(-2 A b+3 a B+b B x) \sqrt {d+e x}}{b^2 \sqrt {a+b x}}+\frac {(b B d+2 A b e-3 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{5/2} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(125)=250\).
time = 0.10, size = 386, normalized size = 2.66

method	result	size
default	\(\frac {\sqrt {e x +d}\, \left (2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} e x -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b e x +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d x +2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b e -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} e +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b d +2 B b x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-4 A b \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+6 B a \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right )}{2 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} \sqrt {b x +a}}\)	\(386\)

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,method=_RETURNVERBOSE)

[Out]

1/2*(e*x+d)^(1/2)*(2*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*e*x-3*B
*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*e*x+B*ln(1/2*(2*b*e*x+2*((b*x
+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d*x+2*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*e-3*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^
(1/2))*a^2*e+B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d+2*B*b*x*((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 1.68, size = 363, normalized size = 2.50 \begin {gather*} \left [\frac {{\left ({\left (B b^{2} d x + B a b d - {\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} e\right )} \sqrt {b} e^{\frac {1}{2}} \log \left (b^{2} d^{2} + 4 \, {\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\frac {1}{2}} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) + 4 \, {\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x e + d} e\right )} e^{\left (-1\right )}}{4 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {{\left (2 \, {\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x e + d} e - {\left (B b^{2} d x + B a b d - {\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} e\right )} \sqrt {-b e} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {-b e} \sqrt {x e + d}}{2 \, {\left ({\left (b^{2} x^{2} + a b x\right )} e^{2} + {\left (b^{2} d x + a b d\right )} e\right )}}\right )\right )} e^{\left (-1\right )}}{2 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \end {gather*}

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

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

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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Giac [A]
time = 0.67, size = 227, normalized size = 1.57 \begin {gather*} \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}} \end {gather*}

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

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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