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

Optimal. Leaf size=174 \[ \frac {3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{7/2}}-\frac {3 \sqrt {a+b x} \sqrt {c+d x} (5 b c-a d)}{4 d^3}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.10, antiderivative size = 174, 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 {(a+b x)^{3/2} \sqrt {c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac {3 \sqrt {a+b x} \sqrt {c+d x} (5 b c-a d)}{4 d^3}+\frac {3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{7/2}}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.42, size = 169, normalized size = 0.97 \begin {gather*} \frac {\frac {\sqrt {d} \left (a^2 d (13 c+5 d x)+a b \left (-15 c^2+8 c d x+7 d^2 x^2\right )+b^2 x \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{\sqrt {a+b x}}+\frac {3 (5 b c-a d) (b c-a d)^{3/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b}}{4 d^{7/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d]*(a^2*d*(13*c + 5*d*x) + b^2*x*(-15*c^2 - 5*c*d*x + 2*d^2*x^2) + a*b*(-15*c^2 + 8*c*d*x + 7*d^2*x^2))
)/Sqrt[a + b*x] + (3*(b*c - a*d)^(3/2)*(5*b*c - a*d)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a +
b*x])/Sqrt[b*c - a*d]])/b)/(4*d^(7/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.87, size = 166, normalized size = 0.95 \begin {gather*} \frac {\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (5 a d (c+d x)+8 a c d-8 b c^2-9 b c (c+d x)+2 b (c+d x)^2\right )}{4 d^3 \sqrt {c+d x}}-\frac {3 \sqrt {\frac {b}{d}} \left (a^2 d^2-6 a b c d+5 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 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - (b*c)/d + (b*(c + d*x))/d]*(-8*b*c^2 + 8*a*c*d - 9*b*c*(c + d*x) + 5*a*d*(c + d*x) + 2*b*(c + d*x)^2
))/(4*d^3*Sqrt[c + d*x]) - (3*Sqrt[b/d]*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Log[-(Sqrt[b/d]*Sqrt[c + d*x]) + Sqr
t[a - (b*c)/d + (b*(c + d*x))/d]])/(4*b*d^3)

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fricas [A]  time = 1.74, size = 434, normalized size = 2.49 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + {\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\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^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b d^{5} x + b c d^{4}\right )}}, -\frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + {\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\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^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b d^{5} x + b c d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.41, size = 206, normalized size = 1.18 \begin {gather*} \frac {\sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} {\left | b \right |}}{b d} - \frac {5 \, b^{2} c d^{3} {\left | b \right |} - a b d^{4} {\left | b \right |}}{b^{2} d^{5}}\right )} - \frac {3 \, {\left (5 \, b^{3} c^{2} d^{2} {\left | b \right |} - 6 \, a b^{2} c d^{3} {\left | b \right |} + a^{2} b d^{4} {\left | b \right |}\right )}}{b^{2} d^{5}}\right )}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {3 \, {\left (5 \, b^{2} c^{2} {\left | b \right |} - 6 \, a b c d {\left | b \right |} + 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 |}\right )}{4 \, \sqrt {b d} b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 455, normalized size = 2.61 \begin {gather*} \frac {\sqrt {b x +a}\, \left (3 a^{2} d^{3} 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 c \,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 )+15 b^{2} c^{2} 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 a^{2} c \,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 b \,c^{2} 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 )+15 b^{2} c^{3} \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 \,d^{2} x^{2}+10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,d^{2} x -10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b c d x +26 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a c d -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{2}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {d x +c}\, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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