3.1505 \(\int \frac {\sqrt {a+b x}}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ \frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {2 \sqrt {a+b x}}{d \sqrt {c+d x}} \]

[Out]

2*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(3/2)-2*(b*x+a)^(1/2)/d/(d*x+c)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {47, 63, 217, 206} \[ \frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {2 \sqrt {a+b x}}{d \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]/(c + d*x)^(3/2),x]

[Out]

(-2*Sqrt[a + b*x])/(d*Sqrt[c + d*x]) + (2*Sqrt[b]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/d^
(3/2)

Rule 47

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

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Mathematica [A]  time = 0.35, size = 95, normalized size = 1.44 \[ \frac {2 \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )-2 \sqrt {d} \sqrt {a+b x}}{d^{3/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 241, normalized size = 3.65 \[ \left [\frac {{\left (d x + c\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (d^{2} x + c d\right )}}, -\frac {{\left (d x + c\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, \sqrt {b x + a} \sqrt {d x + c}}{d^{2} x + c d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((d*x + c)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x + b*c*d + a*d^2)*sq
rt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - 4*sqrt(b*x + a)*sqrt(d*x + c))/(d^2*x + c*d),
-((d*x + c)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a
*b*c + (b^2*c + a*b*d)*x)) + 2*sqrt(b*x + a)*sqrt(d*x + c))/(d^2*x + c*d)]

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giac [A]  time = 1.22, size = 96, normalized size = 1.45 \[ -\frac {2 \, b^{2} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d {\left | b \right |}} - \frac {2 \, \sqrt {b x + a} b^{2}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} d {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^2*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d*abs(b)) - 2*sqrt(
b*x + a)*b^2/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*d*abs(b))

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x +a}}{\left (d x +c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)/(d*x+c)^(3/2),x)

[Out]

int((b*x+a)^(1/2)/(d*x+c)^(3/2),x)

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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)^(1/2)/(d*x+c)^(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*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {a+b\,x}}{{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)/(c + d*x)^(3/2),x)

[Out]

int((a + b*x)^(1/2)/(c + d*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)/(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a + b*x)/(c + d*x)**(3/2), x)

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