3.14.84 \(\int \frac {\sqrt {c+d x}}{(a+b x)^4} \, dx\) [1384]

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

[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 = 0.07, 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 = {43, 44, 65, 214} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x)^4,x]

[Out]

-1/3*Sqrt[c + d*x]/(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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], 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 \text {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 [A]
time = 0.64, size = 131, normalized size = 0.90 \begin {gather*} \frac {\sqrt {c+d x} \left (-3 a^2 d^2+2 a b d (7 c+4 d x)+b^2 \left (-8 c^2-2 c d x+3 d^2 x^2\right )\right )}{24 b (b c-a d)^2 (a+b x)^3}+\frac {d^3 \tan ^{-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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(-3*a^2*d^2 + 2*a*b*d*(7*c + 4*d*x) + b^2*(-8*c^2 - 2*c*d*x + 3*d^2*x^2)))/(24*b*(b*c - a*d)^2*
(a + b*x)^3) + (d^3*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(8*b^(3/2)*(-(b*c) + a*d)^(5/2))

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Maple [A]
time = 0.15, size = 152, normalized size = 1.04

method	result	size
derivativedivides	\(2 d^{3} \left (\frac {\frac {b \left (d x +c \right )^{\frac {5}{2}}}{16 a^{2} d^{2}-32 a b c d +16 b^{2} c^{2}}+\frac {\left (d x +c \right )^{\frac {3}{2}}}{6 a d -6 b c}-\frac {\sqrt {d x +c}}{16 b}}{\left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 b \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )\)	\(152\)
default	\(2 d^{3} \left (\frac {\frac {b \left (d x +c \right )^{\frac {5}{2}}}{16 a^{2} d^{2}-32 a b c d +16 b^{2} c^{2}}+\frac {\left (d x +c \right )^{\frac {3}{2}}}{6 a d -6 b c}-\frac {\sqrt {d x +c}}{16 b}}{\left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 b \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^3*((1/16*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(5/2)+1/6/(a*d-b*c)*(d*x+c)^(3/2)-1/16*(d*x+c)^(1/2)/b)/((d
*x+c)*b+a*d-b*c)^3+1/16/b/(a^2*d^2-2*a*b*c*d+b^2*c^2)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)
^(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((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 detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (122) = 244\).
time = 0.63, size = 785, normalized size = 5.38 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [A]
time = 2.47, size = 207, normalized size = 1.42 \begin {gather*} \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}} \end {gather*}

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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Mupad [B]
time = 0.37, size = 207, normalized size = 1.42 \begin {gather*} \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}} \end {gather*}

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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