3.1384 \(\int \frac {\sqrt {c+d x}}{(a+b x)^4} \, dx\)

Optimal. Leaf size=146 \[ -\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{3/2} (b c-a d)^{5/2}}+\frac {d^2 \sqrt {c+d x}}{8 b (a+b x) (b c-a d)^2}-\frac {d \sqrt {c+d x}}{12 b (a+b x)^2 (b c-a d)}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3} \]

[Out]

-1/8*d^3*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(3/2)/(-a*d+b*c)^(5/2)-1/3*(d*x+c)^(1/2)/b/(b*x+a)^
3-1/12*d*(d*x+c)^(1/2)/b/(-a*d+b*c)/(b*x+a)^2+1/8*d^2*(d*x+c)^(1/2)/b/(-a*d+b*c)^2/(b*x+a)

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Rubi [A]  time = 0.10, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {47, 51, 63, 208} \[ -\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{3/2} (b c-a d)^{5/2}}+\frac {d^2 \sqrt {c+d x}}{8 b (a+b x) (b c-a d)^2}-\frac {d \sqrt {c+d x}}{12 b (a+b x)^2 (b c-a d)}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x)^4,x]

[Out]

-Sqrt[c + d*x]/(3*b*(a + b*x)^3) - (d*Sqrt[c + d*x])/(12*b*(b*c - a*d)*(a + b*x)^2) + (d^2*Sqrt[c + d*x])/(8*b
*(b*c - a*d)^2*(a + b*x)) - (d^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*b^(3/2)*(b*c - a*d)^(5/2
))

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)^4} \, dx &=-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}+\frac {d \int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx}{6 b}\\ &=-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}-\frac {d \sqrt {c+d x}}{12 b (b c-a d) (a+b x)^2}-\frac {d^2 \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{8 b (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}-\frac {d \sqrt {c+d x}}{12 b (b c-a d) (a+b x)^2}+\frac {d^2 \sqrt {c+d x}}{8 b (b c-a d)^2 (a+b x)}+\frac {d^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{16 b (b c-a d)^2}\\ &=-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}-\frac {d \sqrt {c+d x}}{12 b (b c-a d) (a+b x)^2}+\frac {d^2 \sqrt {c+d x}}{8 b (b c-a d)^2 (a+b x)}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{8 b (b c-a d)^2}\\ &=-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}-\frac {d \sqrt {c+d x}}{12 b (b c-a d) (a+b x)^2}+\frac {d^2 \sqrt {c+d x}}{8 b (b c-a d)^2 (a+b x)}-\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{3/2} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.36 \[ \frac {2 d^3 (c+d x)^{3/2} \, _2F_1\left (\frac {3}{2},4;\frac {5}{2};-\frac {b (c+d x)}{a d-b c}\right )}{3 (a d-b c)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]/(a + b*x)^4,x]

[Out]

(2*d^3*(c + d*x)^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^4)

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fricas [B]  time = 0.46, size = 785, normalized size = 5.38 \[ \left [\frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} c^{3} - 22 \, a b^{3} c^{2} d + 17 \, a^{2} b^{2} c d^{2} - 3 \, a^{3} b d^{3} - 3 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (b^{4} c^{2} d - 5 \, a b^{3} c d^{2} + 4 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{48 \, {\left (a^{3} b^{5} c^{3} - 3 \, a^{4} b^{4} c^{2} d + 3 \, a^{5} b^{3} c d^{2} - a^{6} b^{2} d^{3} + {\left (b^{8} c^{3} - 3 \, a b^{7} c^{2} d + 3 \, a^{2} b^{6} c d^{2} - a^{3} b^{5} d^{3}\right )} x^{3} + 3 \, {\left (a b^{7} c^{3} - 3 \, a^{2} b^{6} c^{2} d + 3 \, a^{3} b^{5} c d^{2} - a^{4} b^{4} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{3} - 3 \, a^{3} b^{5} c^{2} d + 3 \, a^{4} b^{4} c d^{2} - a^{5} b^{3} d^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (8 \, b^{4} c^{3} - 22 \, a b^{3} c^{2} d + 17 \, a^{2} b^{2} c d^{2} - 3 \, a^{3} b d^{3} - 3 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (b^{4} c^{2} d - 5 \, a b^{3} c d^{2} + 4 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{24 \, {\left (a^{3} b^{5} c^{3} - 3 \, a^{4} b^{4} c^{2} d + 3 \, a^{5} b^{3} c d^{2} - a^{6} b^{2} d^{3} + {\left (b^{8} c^{3} - 3 \, a b^{7} c^{2} d + 3 \, a^{2} b^{6} c d^{2} - a^{3} b^{5} d^{3}\right )} x^{3} + 3 \, {\left (a b^{7} c^{3} - 3 \, a^{2} b^{6} c^{2} d + 3 \, a^{3} b^{5} c d^{2} - a^{4} b^{4} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{3} - 3 \, a^{3} b^{5} c^{2} d + 3 \, a^{4} b^{4} c d^{2} - a^{5} b^{3} d^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)/(b*x+a)^4,x, algorithm="fricas")

[Out]

[1/48*(3*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*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*(8*b^4*c^3 - 22*a*b^3*c^2*d + 17*a^2*b^2*c*d^2 - 3*a^3
*b*d^3 - 3*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(b^4*c^2*d - 5*a*b^3*c*d^2 + 4*a^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^3*
b^5*c^3 - 3*a^4*b^4*c^2*d + 3*a^5*b^3*c*d^2 - a^6*b^2*d^3 + (b^8*c^3 - 3*a*b^7*c^2*d + 3*a^2*b^6*c*d^2 - a^3*b
^5*d^3)*x^3 + 3*(a*b^7*c^3 - 3*a^2*b^6*c^2*d + 3*a^3*b^5*c*d^2 - a^4*b^4*d^3)*x^2 + 3*(a^2*b^6*c^3 - 3*a^3*b^5
*c^2*d + 3*a^4*b^4*c*d^2 - a^5*b^3*d^3)*x), 1/24*(3*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*
sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (8*b^4*c^3 - 22*a*b^3*c^2*d +
17*a^2*b^2*c*d^2 - 3*a^3*b*d^3 - 3*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(b^4*c^2*d - 5*a*b^3*c*d^2 + 4*a^2*b^2*d^3)
*x)*sqrt(d*x + c))/(a^3*b^5*c^3 - 3*a^4*b^4*c^2*d + 3*a^5*b^3*c*d^2 - a^6*b^2*d^3 + (b^8*c^3 - 3*a*b^7*c^2*d +
3*a^2*b^6*c*d^2 - a^3*b^5*d^3)*x^3 + 3*(a*b^7*c^3 - 3*a^2*b^6*c^2*d + 3*a^3*b^5*c*d^2 - a^4*b^4*d^3)*x^2 + 3*
(a^2*b^6*c^3 - 3*a^3*b^5*c^2*d + 3*a^4*b^4*c*d^2 - a^5*b^3*d^3)*x)]

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giac [A]  time = 1.35, size = 207, normalized size = 1.42 \[ \frac {d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{3} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{3} - 3 \, \sqrt {d x + c} b^{2} c^{2} d^{3} + 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{4} + 6 \, \sqrt {d x + c} a b c d^{4} - 3 \, \sqrt {d x + c} a^{2} d^{5}}{24 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)/(b*x+a)^4,x, algorithm="giac")

[Out]

1/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*sqrt(-b^2*c + a*b*d)
) + 1/24*(3*(d*x + c)^(5/2)*b^2*d^3 - 8*(d*x + c)^(3/2)*b^2*c*d^3 - 3*sqrt(d*x + c)*b^2*c^2*d^3 + 8*(d*x + c)^
(3/2)*a*b*d^4 + 6*sqrt(d*x + c)*a*b*c*d^4 - 3*sqrt(d*x + c)*a^2*d^5)/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*((d*
x + c)*b - b*c + a*d)^3)

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maple [A]  time = 0.02, size = 170, normalized size = 1.16 \[ \frac {\left (d x +c \right )^{\frac {5}{2}} b \,d^{3}}{8 \left (b d x +a d \right )^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, b}+\frac {\left (d x +c \right )^{\frac {3}{2}} d^{3}}{3 \left (b d x +a d \right )^{3} \left (a d -b c \right )}-\frac {\sqrt {d x +c}\, d^{3}}{8 \left (b d x +a d \right )^{3} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(1/2)/(b*x+a)^4,x)

[Out]

1/8*d^3/(b*d*x+a*d)^3*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(5/2)+1/3*d^3/(b*d*x+a*d)^3/(a*d-b*c)*(d*x+c)^(3/2
)-1/8*d^3/(b*d*x+a*d)^3/b*(d*x+c)^(1/2)+1/8*d^3/b/(a^2*d^2-2*a*b*c*d+b^2*c^2)/((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)^4,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 = 0.37, size = 207, normalized size = 1.42 \[ \frac {\frac {d^3\,{\left (c+d\,x\right )}^{3/2}}{3\,\left (a\,d-b\,c\right )}-\frac {d^3\,\sqrt {c+d\,x}}{8\,b}+\frac {b\,d^3\,{\left (c+d\,x\right )}^{5/2}}{8\,{\left (a\,d-b\,c\right )}^2}}{\left (c+d\,x\right )\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )+b^3\,{\left (c+d\,x\right )}^3-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^2+a^3\,d^3-b^3\,c^3+3\,a\,b^2\,c^2\,d-3\,a^2\,b\,c\,d^2}+\frac {d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{8\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/2)/(a + b*x)^4,x)

[Out]

((d^3*(c + d*x)^(3/2))/(3*(a*d - b*c)) - (d^3*(c + d*x)^(1/2))/(8*b) + (b*d^3*(c + d*x)^(5/2))/(8*(a*d - b*c)^
2))/((c + d*x)*(3*b^3*c^2 + 3*a^2*b*d^2 - 6*a*b^2*c*d) + b^3*(c + d*x)^3 - (3*b^3*c - 3*a*b^2*d)*(c + d*x)^2 +
a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2) + (d^3*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/
(8*b^(3/2)*(a*d - b*c)^(5/2))

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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)**4,x)

[Out]

Timed out

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