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

Optimal. Leaf size=171 \[ -\frac {(b c-a d)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{7/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (a d+5 b c)}{8 b d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d} \]

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Rubi [A]  time = 0.10, 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 = {80, 50, 63, 217, 206} \begin {gather*} -\frac {(b c-a d)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{7/2}}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{12 b d^2}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (a d+5 b c)}{8 b d^3}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(5*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b*d^3) - ((5*b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*
x])/(12*b*d^2) + ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*b*d) - ((b*c - a*d)^2*(5*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[
a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(3/2)*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 80

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

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

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Mathematica [A]  time = 0.49, size = 159, normalized size = 0.93 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) \left (3 a^2 d^2+2 a b d (7 d x-11 c)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )-3 (b c-a d)^{5/2} (a d+5 b c) \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 )}{24 b^2 d^{7/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(3*a^2*d^2 + 2*a*b*d*(-11*c + 7*d*x) + b^2*(15*c^2 - 10*c*d*x + 8*d^2*x^2))
- 3*(b*c - a*d)^(5/2)*(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]])/(24*b^2*d^(7/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.36, size = 214, normalized size = 1.25 \begin {gather*} \frac {\sqrt {c+d x} (b c-a d)^2 \left (\frac {15 b^3 c (c+d x)^2}{(a+b x)^2}+\frac {3 a b^2 d (c+d x)^2}{(a+b x)^2}-\frac {40 b^2 c d (c+d x)}{a+b x}-\frac {8 a b d^2 (c+d x)}{a+b x}-3 a d^3+33 b c d^2\right )}{24 b d^3 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^3}-\frac {(b c-a d)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{3/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*Sqrt[c + d*x]*(33*b*c*d^2 - 3*a*d^3 - (40*b^2*c*d*(c + d*x))/(a + b*x) - (8*a*b*d^2*(c + d*x))/
(a + b*x) + (15*b^3*c*(c + d*x)^2)/(a + b*x)^2 + (3*a*b^2*d*(c + d*x)^2)/(a + b*x)^2))/(24*b*d^3*Sqrt[a + b*x]
*(-d + (b*(c + d*x))/(a + b*x))^3) - ((b*c - a*d)^2*(5*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqr
t[a + b*x])])/(8*b^(3/2)*d^(7/2))

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fricas [A]  time = 1.40, size = 412, normalized size = 2.41 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{2} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{2} d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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*(8
*b^3*d^3*x^2 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3)*x)*sqrt(b*x + a)*sq
rt(d*x + c))/(b^2*d^4), 1/48*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d^3*x^2 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^2*d^4)]

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giac [A]  time = 1.49, size = 213, normalized size = 1.25 \begin {gather*} \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2} d} - \frac {5 \, b^{3} c d^{3} + a b^{2} d^{4}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} - a^{2} b^{2} d^{4}\right )}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 )}{\sqrt {b d} b d^{3}}\right )} b}{24 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(b^2*d) - (5*b^3*c*d^3 + a*b
^2*d^4)/(b^4*d^5)) + 3*(5*b^4*c^2*d^2 - 4*a*b^3*c*d^3 - a^2*b^2*d^4)/(b^4*d^5)) + 3*(5*b^3*c^3 - 9*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 + a^3*d^3)*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))*b/abs(b)

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maple [B]  time = 0.02, size = 395, normalized size = 2.31 \begin {gather*} -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 a^{3} d^{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 )+9 a^{2} b 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 )-27 a \,b^{2} 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^{3} 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 )-16 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} d^{2} x^{2}-28 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x +20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+44 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,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)^(1/2),x)

[Out]

-1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-16*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*d^2*x^2+3*a^3*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))+9*a^2*b*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))-27*a*b^2*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^3*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))
-28*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c*d*x-6*(b*d)^(1/
2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+44*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d-30*(b*d)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)*b^2*c^2)/b/d^3/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)

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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)^(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 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}}{\sqrt {c+d\,x}} \,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)^(1/2),x)

[Out]

int((x*(a + b*x)^(3/2))/(c + d*x)^(1/2), x)

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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*(b*x+a)**(3/2)/(d*x+c)**(1/2),x)

[Out]

Timed out

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