3.15.21 \(\int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx\) [1421]

Optimal. Leaf size=114 \[ -\frac {\sqrt {c+d x}}{2 (b c-a d) (a+b x)^2}+\frac {3 d \sqrt {c+d x}}{4 (b c-a d)^2 (a+b x)}-\frac {3 d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 \sqrt {b} (b c-a d)^{5/2}} \]

[Out]

-3/4*d^2*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(5/2)/b^(1/2)-1/2*(d*x+c)^(1/2)/(-a*d+b*c)
/(b*x+a)^2+3/4*d*(d*x+c)^(1/2)/(-a*d+b*c)^2/(b*x+a)

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Rubi [A]
time = 0.03, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {44, 65, 214} \begin {gather*} -\frac {3 d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 \sqrt {b} (b c-a d)^{5/2}}+\frac {3 d \sqrt {c+d x}}{4 (a+b x) (b c-a d)^2}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^3*Sqrt[c + d*x]),x]

[Out]

-1/2*Sqrt[c + d*x]/((b*c - a*d)*(a + b*x)^2) + (3*d*Sqrt[c + d*x])/(4*(b*c - a*d)^2*(a + b*x)) - (3*d^2*ArcTan
h[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*Sqrt[b]*(b*c - a*d)^(5/2))

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

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Mathematica [A]
time = 0.28, size = 96, normalized size = 0.84 \begin {gather*} \frac {1}{4} \left (\frac {\sqrt {c+d x} (-2 b c+5 a d+3 b d x)}{(b c-a d)^2 (a+b x)^2}+\frac {3 d^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} (-b c+a d)^{5/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^3*Sqrt[c + d*x]),x]

[Out]

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

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Maple [A]
time = 0.16, size = 138, normalized size = 1.21

method	result	size
derivativedivides	\(2 d^{2} \left (\frac {\sqrt {d x +c}}{4 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\frac {3 \sqrt {d x +c}}{8 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}}{a d -b c}\right )\)	\(138\)
default	\(2 d^{2} \left (\frac {\sqrt {d x +c}}{4 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\frac {3 \sqrt {d x +c}}{8 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}}{a d -b c}\right )\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (94) = 188\).
time = 0.99, size = 549, normalized size = 4.82 \begin {gather*} \left [\frac {3 \, {\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} - 7 \, a b^{2} c d + 5 \, a^{2} b d^{2} - 3 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{2} + 2 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x\right )}}, \frac {3 \, {\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} - 7 \, a b^{2} c d + 5 \, a^{2} b d^{2} - 3 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{2} + 2 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{3} \sqrt {c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((a + b*x)**3*sqrt(c + d*x)), x)

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Giac [A]
time = 1.49, size = 148, normalized size = 1.30 \begin {gather*} \frac {3 \, d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {3 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} - 5 \, \sqrt {d x + c} b c d^{2} + 5 \, \sqrt {d x + c} a d^{3}}{4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.33, size = 142, normalized size = 1.25 \begin {gather*} \frac {\frac {5\,d^2\,\sqrt {c+d\,x}}{4\,\left (a\,d-b\,c\right )}+\frac {3\,b\,d^2\,{\left (c+d\,x\right )}^{3/2}}{4\,{\left (a\,d-b\,c\right )}^2}}{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}+\frac {3\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{4\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*d^2*(c + d*x)^(1/2))/(4*(a*d - b*c)) + (3*b*d^2*(c + d*x)^(3/2))/(4*(a*d - b*c)^2))/(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) + (3*d^2*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c
)^(1/2)))/(4*b^(1/2)*(a*d - b*c)^(5/2))

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