3.8.36 \(\int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=171 \[ \frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} \sqrt {d}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-5 a d)}{4 b^3}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d)}{2 b^2 (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d)}{2 b^2 (b c-a d)}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-5 a d)}{4 b^3}+\frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} \sqrt {d}}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x)^(3/2))/(a + b*x)^(3/2),x]

[Out]

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 134, normalized size = 0.78 \begin {gather*} \frac {\sqrt {c+d x} \left (\frac {-15 a^2 d+a b (13 c-5 d x)+b^2 x (5 c+2 d x)}{\sqrt {a+b x}}+\frac {3 (b c-5 a d) \sqrt {b c-a d} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(c + d*x)^(3/2))/(a + b*x)^(3/2),x]

[Out]

(Sqrt[c + d*x]*((-15*a^2*d + a*b*(13*c - 5*d*x) + b^2*x*(5*c + 2*d*x))/Sqrt[a + b*x] + (3*(b*c - 5*a*d)*Sqrt[b
*c - a*d]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[(b*(c + d*x))/(b*c - a*d)])))/(4*b^3
)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.93, size = 225, normalized size = 1.32 \begin {gather*} \frac {\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (15 a^2 d^2 \sqrt {c+d x}+5 a b d (c+d x)^{3/2}-18 a b c d \sqrt {c+d x}+3 b^2 c^2 \sqrt {c+d x}-2 b^2 (c+d x)^{5/2}-b^2 c (c+d x)^{3/2}\right )}{4 b^3 (-a d-b (c+d x)+b c)}-\frac {3 \sqrt {\frac {b}{d}} \left (5 a^2 d^2-6 a b c d+b^2 c^2\right ) \log \left (\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(c + d*x)^(3/2))/(a + b*x)^(3/2),x]

[Out]

(Sqrt[a - (b*c)/d + (b*(c + d*x))/d]*(3*b^2*c^2*Sqrt[c + d*x] - 18*a*b*c*d*Sqrt[c + d*x] + 15*a^2*d^2*Sqrt[c +
d*x] - b^2*c*(c + d*x)^(3/2) + 5*a*b*d*(c + d*x)^(3/2) - 2*b^2*(c + d*x)^(5/2)))/(4*b^3*(b*c - a*d - b*(c + d
*x))) - (3*Sqrt[b/d]*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*Log[-(Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[a - (b*c)/d + (b*
(c + d*x))/d]])/(4*b^4)

________________________________________________________________________________________

fricas [A]  time = 1.59, size = 434, normalized size = 2.54 \begin {gather*} \left [\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{5} d x + a b^{4} d\right )}}, -\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{5} d x + a b^{4} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x)*sqrt(b*d)*log(8*b^2*d
^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^
2*c*d + a*b*d^2)*x) + 4*(2*b^3*d^2*x^2 + 13*a*b^2*c*d - 15*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^5*d*x + a*b^4*d), -1/8*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d +
5*a^2*b*d^2)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^
2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 2*(2*b^3*d^2*x^2 + 13*a*b^2*c*d - 15*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^
2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d*x + a*b^4*d)]

________________________________________________________________________________________

giac [A]  time = 1.73, size = 270, normalized size = 1.58 \begin {gather*} \frac {1}{4} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d {\left | b \right |}}{b^{5}} + \frac {5 \, b^{10} c d^{2} {\left | b \right |} - 9 \, a b^{9} d^{3} {\left | b \right |}}{b^{14} d^{2}}\right )} + \frac {4 \, {\left (\sqrt {b d} a b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a^{2} b c d {\left | b \right |} + \sqrt {b d} a^{3} d^{2} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} b^{4}} - \frac {3 \, {\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 6 \, \sqrt {b d} a b c d {\left | b \right |} + 5 \, \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{8 \, b^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d*abs(b)/b^5 + (5*b^10*c*d^2*abs(b) - 9*a*b
^9*d^3*abs(b))/(b^14*d^2)) + 4*(sqrt(b*d)*a*b^2*c^2*abs(b) - 2*sqrt(b*d)*a^2*b*c*d*abs(b) + sqrt(b*d)*a^3*d^2*
abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*b^4) - 3/8*(sqrt(
b*d)*b^2*c^2*abs(b) - 6*sqrt(b*d)*a*b*c*d*abs(b) + 5*sqrt(b*d)*a^2*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^5*d)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 455, normalized size = 2.66 \begin {gather*} \frac {\sqrt {d x +c}\, \left (15 a^{2} b \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-18 a \,b^{2} c d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{3} c^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+15 a^{3} d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-18 a^{2} b c d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a \,b^{2} c^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} d \,x^{2}-10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b d x +10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c x -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} d +26 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b c \right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a^2*b*d^2-
18*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a*b^2*c*d+3*ln(1/2*(2*b*d*x+a
*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*b^3*c^2+4*x^2*b^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)
^(1/2)+15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*d^2-18*ln(1/2*(2*b*d
*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2*b*c*d+3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*
(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a*b^2*c^2-10*x*a*b*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+10*x*b^2*c*(
(b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-30*a^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+26*a*b*c*((b*x+a)*(d*x+c))^(1/
2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^3

________________________________________________________________________________________

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)^(3/2)/(b*x+a)^(3/2),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 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________