3.14.16 \(\int \frac {1}{(a+b x)^4 \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=147 \[ \frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 \sqrt {b} (b c-a d)^{7/2}}-\frac {5 d^2 \sqrt {c+d x}}{8 (a+b x) (b c-a d)^3}+\frac {5 d \sqrt {c+d x}}{12 (a+b x)^2 (b c-a d)^2}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)} \]

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Rubi [A]  time = 0.05, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 208} \begin {gather*} -\frac {5 d^2 \sqrt {c+d x}}{8 (a+b x) (b c-a d)^3}+\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 \sqrt {b} (b c-a d)^{7/2}}+\frac {5 d \sqrt {c+d x}}{12 (a+b x)^2 (b c-a d)^2}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^4*Sqrt[c + d*x]),x]

[Out]

-Sqrt[c + d*x]/(3*(b*c - a*d)*(a + b*x)^3) + (5*d*Sqrt[c + d*x])/(12*(b*c - a*d)^2*(a + b*x)^2) - (5*d^2*Sqrt[
c + d*x])/(8*(b*c - a*d)^3*(a + b*x)) + (5*d^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*Sqrt[b]*(b
*c - a*d)^(7/2))

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 {1}{(a+b x)^4 \sqrt {c+d x}} \, dx &=-\frac {\sqrt {c+d x}}{3 (b c-a d) (a+b x)^3}-\frac {(5 d) \int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx}{6 (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{3 (b c-a d) (a+b x)^3}+\frac {5 d \sqrt {c+d x}}{12 (b c-a d)^2 (a+b x)^2}+\frac {\left (5 d^2\right ) \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{8 (b c-a d)^2}\\ &=-\frac {\sqrt {c+d x}}{3 (b c-a d) (a+b x)^3}+\frac {5 d \sqrt {c+d x}}{12 (b c-a d)^2 (a+b x)^2}-\frac {5 d^2 \sqrt {c+d x}}{8 (b c-a d)^3 (a+b x)}-\frac {\left (5 d^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{16 (b c-a d)^3}\\ &=-\frac {\sqrt {c+d x}}{3 (b c-a d) (a+b x)^3}+\frac {5 d \sqrt {c+d x}}{12 (b c-a d)^2 (a+b x)^2}-\frac {5 d^2 \sqrt {c+d x}}{8 (b c-a d)^3 (a+b x)}-\frac {\left (5 d^2\right ) \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 c-a d)^3}\\ &=-\frac {\sqrt {c+d x}}{3 (b c-a d) (a+b x)^3}+\frac {5 d \sqrt {c+d x}}{12 (b c-a d)^2 (a+b x)^2}-\frac {5 d^2 \sqrt {c+d x}}{8 (b c-a d)^3 (a+b x)}+\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 \sqrt {b} (b c-a d)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.34 \begin {gather*} \frac {2 d^3 \sqrt {c+d x} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};-\frac {b (c+d x)}{a d-b c}\right )}{(a d-b c)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^4*Sqrt[c + d*x]),x]

[Out]

(2*d^3*Sqrt[c + d*x]*Hypergeometric2F1[1/2, 4, 3/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(-(b*c) + a*d)^4

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IntegrateAlgebraic [A]  time = 0.27, size = 173, normalized size = 1.18 \begin {gather*} \frac {d^3 \sqrt {c+d x} \left (33 a^2 d^2+40 a b d (c+d x)-66 a b c d+33 b^2 c^2+15 b^2 (c+d x)^2-40 b^2 c (c+d x)\right )}{24 (b c-a d)^3 (-a d-b (c+d x)+b c)^3}+\frac {5 d^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{8 \sqrt {b} (b c-a d)^3 \sqrt {a d-b c}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^4*Sqrt[c + d*x]),x]

[Out]

(d^3*Sqrt[c + d*x]*(33*b^2*c^2 - 66*a*b*c*d + 33*a^2*d^2 - 40*b^2*c*(c + d*x) + 40*a*b*d*(c + d*x) + 15*b^2*(c
+ d*x)^2))/(24*(b*c - a*d)^3*(b*c - a*d - b*(c + d*x))^3) + (5*d^3*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c
+ d*x])/(b*c - a*d)])/(8*Sqrt[b]*(b*c - a*d)^3*Sqrt[-(b*c) + a*d])

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fricas [B]  time = 1.38, size = 884, normalized size = 6.01 \begin {gather*} \left [-\frac {15 \, {\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} - 34 \, a b^{3} c^{2} d + 59 \, a^{2} b^{2} c d^{2} - 33 \, a^{3} b d^{3} + 15 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} - 10 \, {\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^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{3} + 3 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x\right )}}, -\frac {15 \, {\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} - 34 \, a b^{3} c^{2} d + 59 \, a^{2} b^{2} c d^{2} - 33 \, a^{3} b d^{3} + 15 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} - 10 \, {\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^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{3} + 3 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(15*(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 - 34*a*b^3*c^2*d + 59*a^2*b^2*c*d^2 - 33*
a^3*b*d^3 + 15*(b^4*c*d^2 - a*b^3*d^3)*x^2 - 10*(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^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d +
6*a^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^3 + 3*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 -
4*a^4*b^4*c*d^3 + a^5*b^3*d^4)*x^2 + 3*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 +
a^6*b^2*d^4)*x), -1/24*(15*(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)*arct
an(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (8*b^4*c^3 - 34*a*b^3*c^2*d + 59*a^2*b^2*c*d^2 - 33*a^3
*b*d^3 + 15*(b^4*c*d^2 - a*b^3*d^3)*x^2 - 10*(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^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a
^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^3 + 3*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a
^4*b^4*c*d^3 + a^5*b^3*d^4)*x^2 + 3*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6
*b^2*d^4)*x)]

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giac [A]  time = 1.02, size = 231, normalized size = 1.57 \begin {gather*} -\frac {5 \, d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {15 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{3} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{3} + 33 \, \sqrt {d x + c} b^{2} c^{2} d^{3} + 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{4} - 66 \, \sqrt {d x + c} a b c d^{4} + 33 \, \sqrt {d x + c} a^{2} d^{5}}{24 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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(1/(b*x+a)^4/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-5/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqr
t(-b^2*c + a*b*d)) - 1/24*(15*(d*x + c)^(5/2)*b^2*d^3 - 40*(d*x + c)^(3/2)*b^2*c*d^3 + 33*sqrt(d*x + c)*b^2*c^
2*d^3 + 40*(d*x + c)^(3/2)*a*b*d^4 - 66*sqrt(d*x + c)*a*b*c*d^4 + 33*sqrt(d*x + c)*a^2*d^5)/((b^3*c^3 - 3*a*b^
2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c + a*d)^3)

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maple [A]  time = 0.01, size = 147, normalized size = 1.00 \begin {gather*} \frac {5 d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a d -b c \right )^{3} \sqrt {\left (a d -b c \right ) b}}+\frac {\sqrt {d x +c}\, d^{3}}{3 \left (a d -b c \right ) \left (b d x +a d \right )^{3}}+\frac {5 \sqrt {d x +c}\, d^{3}}{12 \left (a d -b c \right )^{2} \left (b d x +a d \right )^{2}}+\frac {5 \sqrt {d x +c}\, d^{3}}{8 \left (a d -b c \right )^{3} \left (b d x +a d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d^3*(d*x+c)^(1/2)/(a*d-b*c)/(b*d*x+a*d)^3+5/12*d^3/(a*d-b*c)^2*(d*x+c)^(1/2)/(b*d*x+a*d)^2+5/8*d^3/(a*d-b*
c)^3*(d*x+c)^(1/2)/(b*d*x+a*d)+5/8*d^3/(a*d-b*c)^3/((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^4/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.39, size = 218, normalized size = 1.48 \begin {gather*} \frac {\frac {11\,d^3\,\sqrt {c+d\,x}}{8\,\left (a\,d-b\,c\right )}+\frac {5\,b^2\,d^3\,{\left (c+d\,x\right )}^{5/2}}{8\,{\left (a\,d-b\,c\right )}^3}+\frac {5\,b\,d^3\,{\left (c+d\,x\right )}^{3/2}}{3\,{\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 {5\,d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{8\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((11*d^3*(c + d*x)^(1/2))/(8*(a*d - b*c)) + (5*b^2*d^3*(c + d*x)^(5/2))/(8*(a*d - b*c)^3) + (5*b*d^3*(c + d*x)
^(3/2))/(3*(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) + (5*d^3*atan((b^(1/2)*(c + d*x)^(1
/2))/(a*d - b*c)^(1/2)))/(8*b^(1/2)*(a*d - b*c)^(7/2))

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sympy [F(-1)]  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(1/(b*x+a)**4/(d*x+c)**(1/2),x)

[Out]

Timed out

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