3.1383 \(\int \frac {\sqrt {c+d x}}{(a+b x)^3} \, dx\)

Optimal. Leaf size=110 \[ \frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} (b c-a d)^{3/2}}-\frac {d \sqrt {c+d x}}{4 b (a+b x) (b c-a d)}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2} \]

[Out]

1/4*d^2*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(3/2)/(-a*d+b*c)^(3/2)-1/2*(d*x+c)^(1/2)/b/(b*x+a)^2
-1/4*d*(d*x+c)^(1/2)/b/(-a*d+b*c)/(b*x+a)

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Rubi [A]  time = 0.08, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {47, 51, 63, 208} \[ \frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} (b c-a d)^{3/2}}-\frac {d \sqrt {c+d x}}{4 b (a+b x) (b c-a d)}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x)^3,x]

[Out]

-Sqrt[c + d*x]/(2*b*(a + b*x)^2) - (d*Sqrt[c + d*x])/(4*b*(b*c - a*d)*(a + b*x)) + (d^2*ArcTanh[(Sqrt[b]*Sqrt[
c + d*x])/Sqrt[b*c - a*d]])/(4*b^(3/2)*(b*c - a*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {\sqrt {c+d x}}{(a+b x)^3} \, dx &=-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}+\frac {d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{4 b}\\ &=-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}-\frac {d \sqrt {c+d x}}{4 b (b c-a d) (a+b x)}-\frac {d^2 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 b (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}-\frac {d \sqrt {c+d x}}{4 b (b c-a d) (a+b x)}-\frac {d \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 b (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}-\frac {d \sqrt {c+d x}}{4 b (b c-a d) (a+b x)}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.47 \[ \frac {2 d^2 (c+d x)^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};-\frac {b (c+d x)}{a d-b c}\right )}{3 (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*(c + d*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^3)

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fricas [B]  time = 0.46, size = 456, normalized size = 4.15 \[ \left [-\frac {{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (2 \, b^{3} c^{2} - 3 \, a b^{2} c d + a^{2} b d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{4} c^{2} - 2 \, a^{3} b^{3} c d + a^{4} b^{2} d^{2} + {\left (b^{6} c^{2} - 2 \, a b^{5} c d + a^{2} b^{4} d^{2}\right )} x^{2} + 2 \, {\left (a b^{5} c^{2} - 2 \, a^{2} b^{4} c d + a^{3} b^{3} d^{2}\right )} x\right )}}, -\frac {{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) + {\left (2 \, b^{3} c^{2} - 3 \, a b^{2} c d + a^{2} b d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{4} c^{2} - 2 \, a^{3} b^{3} c d + a^{4} b^{2} d^{2} + {\left (b^{6} c^{2} - 2 \, a b^{5} c d + a^{2} b^{4} d^{2}\right )} x^{2} + 2 \, {\left (a b^{5} c^{2} - 2 \, a^{2} b^{4} c d + a^{3} b^{3} d^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b
*d)*sqrt(d*x + c))/(b*x + a)) + 2*(2*b^3*c^2 - 3*a*b^2*c*d + a^2*b*d^2 + (b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x + c
))/(a^2*b^4*c^2 - 2*a^3*b^3*c*d + a^4*b^2*d^2 + (b^6*c^2 - 2*a*b^5*c*d + a^2*b^4*d^2)*x^2 + 2*(a*b^5*c^2 - 2*a
^2*b^4*c*d + a^3*b^3*d^2)*x), -1/4*((b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^
2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (2*b^3*c^2 - 3*a*b^2*c*d + a^2*b*d^2 + (b^3*c*d - a*b^2*d^2)*x)*sq
rt(d*x + c))/(a^2*b^4*c^2 - 2*a^3*b^3*c*d + a^4*b^2*d^2 + (b^6*c^2 - 2*a*b^5*c*d + a^2*b^4*d^2)*x^2 + 2*(a*b^5
*c^2 - 2*a^2*b^4*c*d + a^3*b^3*d^2)*x)]

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giac [A]  time = 1.33, size = 126, normalized size = 1.15 \[ -\frac {d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}} - \frac {{\left (d x + c\right )}^{\frac {3}{2}} b d^{2} + \sqrt {d x + c} b c d^{2} - \sqrt {d x + c} a d^{3}}{4 \, {\left (b^{2} c - a b d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^2*c - a*b*d)*sqrt(-b^2*c + a*b*d)) - 1/4*((d*x + c)^
(3/2)*b*d^2 + sqrt(d*x + c)*b*c*d^2 - sqrt(d*x + c)*a*d^3)/((b^2*c - a*b*d)*((d*x + c)*b - b*c + a*d)^2)

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maple [A]  time = 0.01, size = 111, normalized size = 1.01 \[ \frac {d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b}+\frac {\left (d x +c \right )^{\frac {3}{2}} d^{2}}{4 \left (b d x +a d \right )^{2} \left (a d -b c \right )}-\frac {\sqrt {d x +c}\, d^{2}}{4 \left (b d x +a d \right )^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^2/(b*d*x+a*d)^2/(a*d-b*c)*(d*x+c)^(3/2)-1/4*d^2/(b*d*x+a*d)^2/b*(d*x+c)^(1/2)+1/4*d^2/(a*d-b*c)/b/((a*d-
b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)

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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((d*x+c)^(1/2)/(b*x+a)^3,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 positive or negative?

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mupad [B]  time = 0.30, size = 135, normalized size = 1.23 \[ \frac {d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{4\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {\frac {d^2\,\sqrt {c+d\,x}}{4\,b}-\frac {d^2\,{\left (c+d\,x\right )}^{3/2}}{4\,\left (a\,d-b\,c\right )}}{b^2\,{\left (c+d\,x\right )}^2-\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (c+d\,x\right )+a^2\,d^2+b^2\,c^2-2\,a\,b\,c\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/2)/(a + b*x)^3,x)

[Out]

(d^2*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(4*b^(3/2)*(a*d - b*c)^(3/2)) - ((d^2*(c + d*x)^(1/2))
/(4*b) - (d^2*(c + d*x)^(3/2))/(4*(a*d - b*c)))/(b^2*(c + d*x)^2 - (2*b^2*c - 2*a*b*d)*(c + d*x) + a^2*d^2 + b
^2*c^2 - 2*a*b*c*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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