3.5.45 \(\int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=178 \[ -\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}}+\frac {\sqrt {c+d x} (2 b c-7 a d) (b c-a d)}{b^4}+\frac {(c+d x)^{3/2} (2 b c-7 a d)}{3 b^3}+\frac {(c+d x)^{5/2} (2 b c-7 a d)}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)} \]

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Rubi [A]  time = 0.10, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 50, 63, 208} \begin {gather*} \frac {(c+d x)^{5/2} (2 b c-7 a d)}{5 b^2 (b c-a d)}+\frac {(c+d x)^{3/2} (2 b c-7 a d)}{3 b^3}+\frac {\sqrt {c+d x} (2 b c-7 a d) (b c-a d)}{b^4}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x)^(5/2))/(a + b*x)^2,x]

[Out]

((2*b*c - 7*a*d)*(b*c - a*d)*Sqrt[c + d*x])/b^4 + ((2*b*c - 7*a*d)*(c + d*x)^(3/2))/(3*b^3) + ((2*b*c - 7*a*d)
*(c + d*x)^(5/2))/(5*b^2*(b*c - a*d)) + (a*(c + d*x)^(7/2))/(b*(b*c - a*d)*(a + b*x)) - ((2*b*c - 7*a*d)*(b*c
- a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/b^(9/2)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 {x (c+d x)^{5/2}}{(a+b x)^2} \, dx &=\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac {(2 b c-7 a d) \int \frac {(c+d x)^{5/2}}{a+b x} \, dx}{2 b (b c-a d)}\\ &=\frac {(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac {(2 b c-7 a d) \int \frac {(c+d x)^{3/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac {(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac {(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac {((2 b c-7 a d) (b c-a d)) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{2 b^3}\\ &=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x}}{b^4}+\frac {(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac {(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac {\left ((2 b c-7 a d) (b c-a d)^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 b^4}\\ &=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x}}{b^4}+\frac {(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac {(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}+\frac {\left ((2 b c-7 a d) (b c-a 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 )}{b^4 d}\\ &=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x}}{b^4}+\frac {(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac {(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 150, normalized size = 0.84 \begin {gather*} \frac {\frac {2 \left (b c-\frac {7 a d}{2}\right ) \left (5 (b c-a d) \left (\sqrt {b} \sqrt {c+d x} (-3 a d+4 b c+b d x)-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )\right )+3 b^{5/2} (c+d x)^{5/2}\right )}{15 b^{7/2}}+\frac {a (c+d x)^{7/2}}{a+b x}}{b (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(c + d*x)^(5/2))/(a + b*x)^2,x]

[Out]

((a*(c + d*x)^(7/2))/(a + b*x) + (2*(b*c - (7*a*d)/2)*(3*b^(5/2)*(c + d*x)^(5/2) + 5*(b*c - a*d)*(Sqrt[b]*Sqrt
[c + d*x]*(4*b*c - 3*a*d + b*d*x) - 3*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])))/(1
5*b^(7/2)))/(b*(b*c - a*d))

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IntegrateAlgebraic [A]  time = 0.42, size = 229, normalized size = 1.29 \begin {gather*} \frac {\sqrt {a d-b c} \left (7 a^2 d^2-9 a b c d+2 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{b^{9/2}}+\frac {\sqrt {c+d x} \left (105 a^3 d^3+70 a^2 b d^2 (c+d x)-240 a^2 b c d^2+165 a b^2 c^2 d-14 a b^2 d (c+d x)^2-90 a b^2 c d (c+d x)-30 b^3 c^3+20 b^3 c^2 (c+d x)+6 b^3 (c+d x)^3+4 b^3 c (c+d x)^2\right )}{15 b^4 (a d+b (c+d x)-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(c + d*x)^(5/2))/(a + b*x)^2,x]

[Out]

(Sqrt[c + d*x]*(-30*b^3*c^3 + 165*a*b^2*c^2*d - 240*a^2*b*c*d^2 + 105*a^3*d^3 + 20*b^3*c^2*(c + d*x) - 90*a*b^
2*c*d*(c + d*x) + 70*a^2*b*d^2*(c + d*x) + 4*b^3*c*(c + d*x)^2 - 14*a*b^2*d*(c + d*x)^2 + 6*b^3*(c + d*x)^3))/
(15*b^4*(-(b*c) + a*d + b*(c + d*x))) + (Sqrt[-(b*c) + a*d]*(2*b^2*c^2 - 9*a*b*c*d + 7*a^2*d^2)*ArcTan[(Sqrt[b
]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/b^(9/2)

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fricas [A]  time = 1.46, size = 450, normalized size = 2.53 \begin {gather*} \left [\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (6 \, b^{3} d^{2} x^{3} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x\right )} \sqrt {d x + c}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (6 \, b^{3} d^{2} x^{3} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*b^2*c*d + 7*a^2*b*d^2)*x)*sqrt((b*c - a*d)
/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(6*b^3*d^2*x^3 + 61*a*b^2
*c^2 - 170*a^2*b*c*d + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^2 + 2*(23*b^3*c^2 - 59*a*b^2*c*d + 35*a^2*
b*d^2)*x)*sqrt(d*x + c))/(b^5*x + a*b^4), -1/15*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*
b^2*c*d + 7*a^2*b*d^2)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (6*
b^3*d^2*x^3 + 61*a*b^2*c^2 - 170*a^2*b*c*d + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^2 + 2*(23*b^3*c^2 -
59*a*b^2*c*d + 35*a^2*b*d^2)*x)*sqrt(d*x + c))/(b^5*x + a*b^4)]

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giac [A]  time = 1.31, size = 240, normalized size = 1.35 \begin {gather*} \frac {{\left (2 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 16 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{4}} + \frac {\sqrt {d x + c} a b^{2} c^{2} d - 2 \, \sqrt {d x + c} a^{2} b c d^{2} + \sqrt {d x + c} a^{3} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{8} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{8} c + 15 \, \sqrt {d x + c} b^{8} c^{2} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{7} d - 60 \, \sqrt {d x + c} a b^{7} c d + 45 \, \sqrt {d x + c} a^{2} b^{6} d^{2}\right )}}{15 \, b^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*c^3 - 11*a*b^2*c^2*d + 16*a^2*b*c*d^2 - 7*a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-
b^2*c + a*b*d)*b^4) + (sqrt(d*x + c)*a*b^2*c^2*d - 2*sqrt(d*x + c)*a^2*b*c*d^2 + sqrt(d*x + c)*a^3*d^3)/(((d*x
+ c)*b - b*c + a*d)*b^4) + 2/15*(3*(d*x + c)^(5/2)*b^8 + 5*(d*x + c)^(3/2)*b^8*c + 15*sqrt(d*x + c)*b^8*c^2 -
10*(d*x + c)^(3/2)*a*b^7*d - 60*sqrt(d*x + c)*a*b^7*c*d + 45*sqrt(d*x + c)*a^2*b^6*d^2)/b^10

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maple [B]  time = 0.02, size = 348, normalized size = 1.96 \begin {gather*} -\frac {7 a^{3} d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{4}}+\frac {16 a^{2} c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{3}}-\frac {11 a \,c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}+\frac {2 c^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}+\frac {\sqrt {d x +c}\, a^{3} d^{3}}{\left (b d x +a d \right ) b^{4}}-\frac {2 \sqrt {d x +c}\, a^{2} c \,d^{2}}{\left (b d x +a d \right ) b^{3}}+\frac {\sqrt {d x +c}\, a \,c^{2} d}{\left (b d x +a d \right ) b^{2}}+\frac {6 \sqrt {d x +c}\, a^{2} d^{2}}{b^{4}}-\frac {8 \sqrt {d x +c}\, a c d}{b^{3}}+\frac {2 \sqrt {d x +c}\, c^{2}}{b^{2}}-\frac {4 \left (d x +c \right )^{\frac {3}{2}} a d}{3 b^{3}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} c}{3 b^{2}}+\frac {2 \left (d x +c \right )^{\frac {5}{2}}}{5 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/b^2*(d*x+c)^(5/2)-4/3/b^3*(d*x+c)^(3/2)*a*d+2/3/b^2*(d*x+c)^(3/2)*c+6/b^4*(d*x+c)^(1/2)*a^2*d^2-8/b^3*(d*x
+c)^(1/2)*a*c*d+2/b^2*(d*x+c)^(1/2)*c^2+1/b^4*(d*x+c)^(1/2)/(b*d*x+a*d)*a^3*d^3-2/b^3*(d*x+c)^(1/2)/(b*d*x+a*d
)*a^2*c*d^2+1/b^2*(d*x+c)^(1/2)/(b*d*x+a*d)*a*c^2*d-7/b^4/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*
b)^(1/2)*b)*a^3*d^3+16/b^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a^2*c*d^2-11/b^2/((
a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a*c^2*d+2/b/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(
1/2)/((a*d-b*c)*b)^(1/2)*b)*c^3

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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(x*(d*x+c)^(5/2)/(b*x+a)^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 = 0.45, size = 264, normalized size = 1.48 \begin {gather*} \frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b^2}-\left (\frac {2\,{\left (a\,d-b\,c\right )}^2}{b^4}+\frac {\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (\frac {2\,c}{b^2}-\frac {2\,\left (2\,b^2\,c-2\,a\,b\,d\right )}{b^4}\right )}{b^2}\right )\,\sqrt {c+d\,x}-\left (\frac {2\,c}{3\,b^2}-\frac {2\,\left (2\,b^2\,c-2\,a\,b\,d\right )}{3\,b^4}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {\sqrt {c+d\,x}\,\left (a^3\,d^3-2\,a^2\,b\,c\,d^2+a\,b^2\,c^2\,d\right )}{b^5\,\left (c+d\,x\right )-b^5\,c+a\,b^4\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )\,\sqrt {c+d\,x}}{7\,a^3\,d^3-16\,a^2\,b\,c\,d^2+11\,a\,b^2\,c^2\,d-2\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )}{b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(c + d*x)^(5/2))/(5*b^2) - ((2*(a*d - b*c)^2)/b^4 + ((2*b^2*c - 2*a*b*d)*((2*c)/b^2 - (2*(2*b^2*c - 2*a*b*d
))/b^4))/b^2)*(c + d*x)^(1/2) - ((2*c)/(3*b^2) - (2*(2*b^2*c - 2*a*b*d))/(3*b^4))*(c + d*x)^(3/2) + ((c + d*x)
^(1/2)*(a^3*d^3 + a*b^2*c^2*d - 2*a^2*b*c*d^2))/(b^5*(c + d*x) - b^5*c + a*b^4*d) - (atan((b^(1/2)*(a*d - b*c)
^(3/2)*(7*a*d - 2*b*c)*(c + d*x)^(1/2))/(7*a^3*d^3 - 2*b^3*c^3 + 11*a*b^2*c^2*d - 16*a^2*b*c*d^2))*(a*d - b*c)
^(3/2)*(7*a*d - 2*b*c))/b^(9/2)

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

[Out]

Timed out

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