3.1423 \(\int \frac {1}{(a+b x)^5 \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=180 \[ -\frac {35 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 \sqrt {b} (b c-a d)^{9/2}}+\frac {35 d^3 \sqrt {c+d x}}{64 (a+b x) (b c-a d)^4}-\frac {35 d^2 \sqrt {c+d x}}{96 (a+b x)^2 (b c-a d)^3}+\frac {7 d \sqrt {c+d x}}{24 (a+b x)^3 (b c-a d)^2}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)} \]

[Out]

-35/64*d^4*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(9/2)/b^(1/2)-1/4*(d*x+c)^(1/2)/(-a*d+b*
c)/(b*x+a)^4+7/24*d*(d*x+c)^(1/2)/(-a*d+b*c)^2/(b*x+a)^3-35/96*d^2*(d*x+c)^(1/2)/(-a*d+b*c)^3/(b*x+a)^2+35/64*
d^3*(d*x+c)^(1/2)/(-a*d+b*c)^4/(b*x+a)

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Rubi [A]  time = 0.06, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ \frac {35 d^3 \sqrt {c+d x}}{64 (a+b x) (b c-a d)^4}-\frac {35 d^2 \sqrt {c+d x}}{96 (a+b x)^2 (b c-a d)^3}-\frac {35 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 \sqrt {b} (b c-a d)^{9/2}}+\frac {7 d \sqrt {c+d x}}{24 (a+b x)^3 (b c-a d)^2}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^5*Sqrt[c + d*x]),x]

[Out]

-Sqrt[c + d*x]/(4*(b*c - a*d)*(a + b*x)^4) + (7*d*Sqrt[c + d*x])/(24*(b*c - a*d)^2*(a + b*x)^3) - (35*d^2*Sqrt
[c + d*x])/(96*(b*c - a*d)^3*(a + b*x)^2) + (35*d^3*Sqrt[c + d*x])/(64*(b*c - a*d)^4*(a + b*x)) - (35*d^4*ArcT
anh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*Sqrt[b]*(b*c - a*d)^(9/2))

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)^5 \sqrt {c+d x}} \, dx &=-\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}-\frac {(7 d) \int \frac {1}{(a+b x)^4 \sqrt {c+d x}} \, dx}{8 (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}+\frac {7 d \sqrt {c+d x}}{24 (b c-a d)^2 (a+b x)^3}+\frac {\left (35 d^2\right ) \int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx}{48 (b c-a d)^2}\\ &=-\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}+\frac {7 d \sqrt {c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac {35 d^2 \sqrt {c+d x}}{96 (b c-a d)^3 (a+b x)^2}-\frac {\left (35 d^3\right ) \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{64 (b c-a d)^3}\\ &=-\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}+\frac {7 d \sqrt {c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac {35 d^2 \sqrt {c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac {35 d^3 \sqrt {c+d x}}{64 (b c-a d)^4 (a+b x)}+\frac {\left (35 d^4\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{128 (b c-a d)^4}\\ &=-\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}+\frac {7 d \sqrt {c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac {35 d^2 \sqrt {c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac {35 d^3 \sqrt {c+d x}}{64 (b c-a d)^4 (a+b x)}+\frac {\left (35 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{64 (b c-a d)^4}\\ &=-\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}+\frac {7 d \sqrt {c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac {35 d^2 \sqrt {c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac {35 d^3 \sqrt {c+d x}}{64 (b c-a d)^4 (a+b x)}-\frac {35 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 \sqrt {b} (b c-a d)^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.28 \[ \frac {2 d^4 \sqrt {c+d x} \, _2F_1\left (\frac {1}{2},5;\frac {3}{2};-\frac {b (c+d x)}{a d-b c}\right )}{(a d-b c)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^5*Sqrt[c + d*x]),x]

[Out]

(2*d^4*Sqrt[c + d*x]*Hypergeometric2F1[1/2, 5, 3/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(-(b*c) + a*d)^5

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fricas [B]  time = 0.48, size = 1325, normalized size = 7.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^5/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(105*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*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*(48*b^5*c^4 - 248*a*b^4*c^3*d +
526*a^2*b^3*c^2*d^2 - 605*a^3*b^2*c*d^3 + 279*a^4*b*d^4 - 105*(b^5*c*d^3 - a*b^4*d^4)*x^3 + 35*(2*b^5*c^2*d^2
- 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 - 7*(8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2*d
^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^5 - 5*a^5*b^5*c^4*d + 10*a^6*b^4*c^3*d^2 - 10*a^7*b^3*c^2*d^3 + 5*a^8*b^2*c*d
^4 - a^9*b*d^5 + (b^10*c^5 - 5*a*b^9*c^4*d + 10*a^2*b^8*c^3*d^2 - 10*a^3*b^7*c^2*d^3 + 5*a^4*b^6*c*d^4 - a^5*b
^5*d^5)*x^4 + 4*(a*b^9*c^5 - 5*a^2*b^8*c^4*d + 10*a^3*b^7*c^3*d^2 - 10*a^4*b^6*c^2*d^3 + 5*a^5*b^5*c*d^4 - a^6
*b^4*d^5)*x^3 + 6*(a^2*b^8*c^5 - 5*a^3*b^7*c^4*d + 10*a^4*b^6*c^3*d^2 - 10*a^5*b^5*c^2*d^3 + 5*a^6*b^4*c*d^4 -
a^7*b^3*d^5)*x^2 + 4*(a^3*b^7*c^5 - 5*a^4*b^6*c^4*d + 10*a^5*b^5*c^3*d^2 - 10*a^6*b^4*c^2*d^3 + 5*a^7*b^3*c*d
^4 - a^8*b^2*d^5)*x), 1/192*(105*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)
*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (48*b^5*c^4 - 248*a*b^4*c^3*d
+ 526*a^2*b^3*c^2*d^2 - 605*a^3*b^2*c*d^3 + 279*a^4*b*d^4 - 105*(b^5*c*d^3 - a*b^4*d^4)*x^3 + 35*(2*b^5*c^2*d
^2 - 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 - 7*(8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2
*d^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^5 - 5*a^5*b^5*c^4*d + 10*a^6*b^4*c^3*d^2 - 10*a^7*b^3*c^2*d^3 + 5*a^8*b^2*c
*d^4 - a^9*b*d^5 + (b^10*c^5 - 5*a*b^9*c^4*d + 10*a^2*b^8*c^3*d^2 - 10*a^3*b^7*c^2*d^3 + 5*a^4*b^6*c*d^4 - a^5
*b^5*d^5)*x^4 + 4*(a*b^9*c^5 - 5*a^2*b^8*c^4*d + 10*a^3*b^7*c^3*d^2 - 10*a^4*b^6*c^2*d^3 + 5*a^5*b^5*c*d^4 - a
^6*b^4*d^5)*x^3 + 6*(a^2*b^8*c^5 - 5*a^3*b^7*c^4*d + 10*a^4*b^6*c^3*d^2 - 10*a^5*b^5*c^2*d^3 + 5*a^6*b^4*c*d^4
- a^7*b^3*d^5)*x^2 + 4*(a^3*b^7*c^5 - 5*a^4*b^6*c^4*d + 10*a^5*b^5*c^3*d^2 - 10*a^6*b^4*c^2*d^3 + 5*a^7*b^3*c
*d^4 - a^8*b^2*d^5)*x)]

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giac [B]  time = 1.13, size = 331, normalized size = 1.84 \[ \frac {35 \, d^{4} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{64 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b^{2} c + a b d}} + \frac {105 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d^{4} - 385 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} c d^{4} + 511 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} - 279 \, \sqrt {d x + c} b^{3} c^{3} d^{4} + 385 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{2} d^{5} - 1022 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} c d^{5} + 837 \, \sqrt {d x + c} a b^{2} c^{2} d^{5} + 511 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b d^{6} - 837 \, \sqrt {d x + c} a^{2} b c d^{6} + 279 \, \sqrt {d x + c} a^{3} d^{7}}{192 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^5/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

35/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b
*c*d^3 + a^4*d^4)*sqrt(-b^2*c + a*b*d)) + 1/192*(105*(d*x + c)^(7/2)*b^3*d^4 - 385*(d*x + c)^(5/2)*b^3*c*d^4 +
511*(d*x + c)^(3/2)*b^3*c^2*d^4 - 279*sqrt(d*x + c)*b^3*c^3*d^4 + 385*(d*x + c)^(5/2)*a*b^2*d^5 - 1022*(d*x +
c)^(3/2)*a*b^2*c*d^5 + 837*sqrt(d*x + c)*a*b^2*c^2*d^5 + 511*(d*x + c)^(3/2)*a^2*b*d^6 - 837*sqrt(d*x + c)*a^
2*b*c*d^6 + 279*sqrt(d*x + c)*a^3*d^7)/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4
)*((d*x + c)*b - b*c + a*d)^4)

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maple [A]  time = 0.01, size = 179, normalized size = 0.99 \[ \frac {35 d^{4} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{64 \left (a d -b c \right )^{4} \sqrt {\left (a d -b c \right ) b}}+\frac {\sqrt {d x +c}\, d^{4}}{4 \left (a d -b c \right ) \left (b d x +a d \right )^{4}}+\frac {7 \sqrt {d x +c}\, d^{4}}{24 \left (a d -b c \right )^{2} \left (b d x +a d \right )^{3}}+\frac {35 \sqrt {d x +c}\, d^{4}}{96 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}+\frac {35 \sqrt {d x +c}\, d^{4}}{64 \left (a d -b c \right )^{4} \left (b d x +a d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^5/(d*x+c)^(1/2),x)

[Out]

1/4*d^4*(d*x+c)^(1/2)/(a*d-b*c)/(b*d*x+a*d)^4+7/24*d^4/(a*d-b*c)^2*(d*x+c)^(1/2)/(b*d*x+a*d)^3+35/96*d^4/(a*d-
b*c)^3*(d*x+c)^(1/2)/(b*d*x+a*d)^2+35/64*d^4/(a*d-b*c)^4*(d*x+c)^(1/2)/(b*d*x+a*d)+35/64*d^4/(a*d-b*c)^4/((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(1/(b*x+a)^5/(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 = 0.46, size = 307, normalized size = 1.71 \[ \frac {\frac {93\,d^4\,\sqrt {c+d\,x}}{64\,\left (a\,d-b\,c\right )}+\frac {385\,b^2\,d^4\,{\left (c+d\,x\right )}^{5/2}}{192\,{\left (a\,d-b\,c\right )}^3}+\frac {35\,b^3\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,{\left (a\,d-b\,c\right )}^4}+\frac {511\,b\,d^4\,{\left (c+d\,x\right )}^{3/2}}{192\,{\left (a\,d-b\,c\right )}^2}}{b^4\,{\left (c+d\,x\right )}^4-\left (4\,b^4\,c-4\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^3-\left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )+a^4\,d^4+b^4\,c^4+{\left (c+d\,x\right )}^2\,\left (6\,a^2\,b^2\,d^2-12\,a\,b^3\,c\,d+6\,b^4\,c^2\right )+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d-4\,a^3\,b\,c\,d^3}+\frac {35\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^5*(c + d*x)^(1/2)),x)

[Out]

((93*d^4*(c + d*x)^(1/2))/(64*(a*d - b*c)) + (385*b^2*d^4*(c + d*x)^(5/2))/(192*(a*d - b*c)^3) + (35*b^3*d^4*(
c + d*x)^(7/2))/(64*(a*d - b*c)^4) + (511*b*d^4*(c + d*x)^(3/2))/(192*(a*d - b*c)^2))/(b^4*(c + d*x)^4 - (4*b^
4*c - 4*a*b^3*d)*(c + d*x)^3 - (c + d*x)*(4*b^4*c^3 - 4*a^3*b*d^3 + 12*a^2*b^2*c*d^2 - 12*a*b^3*c^2*d) + a^4*d
^4 + b^4*c^4 + (c + d*x)^2*(6*b^4*c^2 + 6*a^2*b^2*d^2 - 12*a*b^3*c*d) + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*
a^3*b*c*d^3) + (35*d^4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(1/2)*(a*d - b*c)^(9/2))

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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(1/(b*x+a)**5/(d*x+c)**(1/2),x)

[Out]

Timed out

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