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

Optimal. Leaf size=169 \[ -\frac {(a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{7/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{24 b^3 d^3}+\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d} \]

[Out]

-1/8*(a*d+b*c)*(5*a^2*d^2-2*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^
(7/2)+1/3*x^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d+1/24*(15*b^2*c^2+14*a*b*c*d+15*a^2*d^2-10*b*d*(a*d+b*c)*x)*(b*x+
a)^(1/2)*(d*x+c)^(1/2)/b^3/d^3

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Rubi [A]  time = 0.11, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {100, 147, 63, 217, 206} \[ \frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{24 b^3 d^3}-\frac {(a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{7/2}}+\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(3*b*d) + (Sqrt[a + b*x]*Sqrt[c + d*x]*(15*b^2*c^2 + 14*a*b*c*d + 15*a^2*d^2
- 10*b*d*(b*c + a*d)*x))/(24*b^3*d^3) - ((b*c + a*d)*(5*b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(7/2)*d^(7/2))

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 100

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx &=\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}+\frac {\int \frac {x \left (-2 a c-\frac {5}{2} (b c+a d) x\right )}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 b d}\\ &=\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 b^2 c^2+14 a b c d+15 a^2 d^2-10 b d (b c+a d) x\right )}{24 b^3 d^3}-\frac {\left ((b c+a d) \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b^3 d^3}\\ &=\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 b^2 c^2+14 a b c d+15 a^2 d^2-10 b d (b c+a d) x\right )}{24 b^3 d^3}-\frac {\left ((b c+a d) \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right )\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^4 d^3}\\ &=\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 b^2 c^2+14 a b c d+15 a^2 d^2-10 b d (b c+a d) x\right )}{24 b^3 d^3}-\frac {\left ((b c+a d) \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right )\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^4 d^3}\\ &=\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 b^2 c^2+14 a b c d+15 a^2 d^2-10 b d (b c+a d) x\right )}{24 b^3 d^3}-\frac {(b c+a d) \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 188, normalized size = 1.11 \[ \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) \left (15 a^2 d^2+2 a b d (7 c-5 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )-3 \sqrt {b c-a d} \left (5 a^3 d^3+3 a^2 b c d^2+3 a b^2 c^2 d+5 b^3 c^3\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{24 b^4 d^{7/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(15*a^2*d^2 + 2*a*b*d*(7*c - 5*d*x) + b^2*(15*c^2 - 10*c*d*x + 8*d^2*x^2))
- 3*Sqrt[b*c - a*d]*(5*b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Ar
cSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(24*b^4*d^(7/2)*Sqrt[c + d*x])

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fricas [A]  time = 1.13, size = 410, normalized size = 2.43 \[ \left [\frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} - 10 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} - 10 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(5*b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*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 + 14*a*b^2*c*d^2 + 15*a^2*b*d^3 - 10*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sq
rt(d*x + c))/(b^4*d^4), 1/48*(3*(5*b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*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 + 14*a*b^2*c*d^2 + 15*a^2*b*d^3 - 10*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^4*d^4)]

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giac [A]  time = 1.17, size = 215, normalized size = 1.27 \[ \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{4} d} - \frac {5 \, b^{12} c d^{3} + 13 \, a b^{11} d^{4}}{b^{15} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{13} c^{2} d^{2} + 8 \, a b^{12} c d^{3} + 11 \, a^{2} b^{11} d^{4}\right )}}{b^{15} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{3} d^{3}}\right )} b}{24 \, {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x+a)^(1/2)/(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^4*d) - (5*b^12*c*d^3 + 13
*a*b^11*d^4)/(b^15*d^5)) + 3*(5*b^13*c^2*d^2 + 8*a*b^12*c*d^3 + 11*a^2*b^11*d^4)/(b^15*d^5)) + 3*(5*b^3*c^3 +
3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*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^3*d^3))*b/abs(b)

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maple [B]  time = 0.03, size = 395, normalized size = 2.34 \[ -\frac {\left (15 a^{3} d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+9 a^{2} b c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+9 a \,b^{2} c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+15 b^{3} c^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-16 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} d^{2} x^{2}+20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x +20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}-28 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right ) \sqrt {b x +a}\, \sqrt {d x +c}}{48 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(-16*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*d^2*x^2+15*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))+9*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))+20*(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-30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*
a^2*d^2-28*(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*x+a)
^(1/2)*(d*x+c)^(1/2)/d^3/b^3/(b*d)^(1/2)/((b*x+a)*(d*x+c))^(1/2)

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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(x^3/(b*x+a)^(1/2)/(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 [B]  time = 41.73, size = 900, normalized size = 5.33 \[ -\frac {\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {85\,a^3\,b\,d^3}{12}+\frac {17\,a^2\,b^2\,c\,d^2}{4}+\frac {17\,a\,b^3\,c^2\,d}{4}+\frac {85\,b^4\,c^3}{12}\right )}{d^8\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {5\,a^3\,b^2\,d^3}{4}+\frac {3\,a^2\,b^3\,c\,d^2}{4}+\frac {3\,a\,b^4\,c^2\,d}{4}+\frac {5\,b^5\,c^3}{4}\right )}{d^9\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {33\,a^3\,d^3}{2}+\frac {327\,a^2\,b\,c\,d^2}{2}+\frac {327\,a\,b^2\,c^2\,d}{2}+\frac {33\,b^3\,c^3}{2}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {64\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (\frac {5\,a^3\,d^3}{4}+\frac {3\,a^2\,b\,c\,d^2}{4}+\frac {3\,a\,b^2\,c^2\,d}{4}+\frac {5\,b^3\,c^3}{4}\right )}{b^3\,d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (\frac {85\,a^3\,d^3}{12}+\frac {17\,a^2\,b\,c\,d^2}{4}+\frac {17\,a\,b^2\,c^2\,d}{4}+\frac {85\,b^3\,c^3}{12}\right )}{b^2\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {33\,a^3\,d^3}{2}+\frac {327\,a^2\,b\,c\,d^2}{2}+\frac {327\,a\,b^2\,c^2\,d}{2}+\frac {33\,b^3\,c^3}{2}\right )}{b\,d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (128\,a^2\,d^2+\frac {896\,a\,b\,c\,d}{3}+128\,b^2\,c^2\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {64\,a^{3/2}\,b^2\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{12}}+\frac {b^6}{d^6}-\frac {6\,b^5\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {15\,b^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {20\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {15\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {6\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (a\,d+b\,c\right )\,\left (5\,a^2\,d^2-2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{4\,b^{7/2}\,d^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((((a + b*x)^(1/2) - a^(1/2))^3*((85*b^4*c^3)/12 + (85*a^3*b*d^3)/12 + (17*a^2*b^2*c*d^2)/4 + (17*a*b^3*c^2*
d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2))^3) - (((a + b*x)^(1/2) - a^(1/2))*((5*b^5*c^3)/4 + (5*a^3*b^2*d^3)/4 +
(3*a^2*b^3*c*d^2)/4 + (3*a*b^4*c^2*d)/4))/(d^9*((c + d*x)^(1/2) - c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^5*
((33*a^3*d^3)/2 + (33*b^3*c^3)/2 + (327*a*b^2*c^2*d)/2 + (327*a^2*b*c*d^2)/2))/(d^7*((c + d*x)^(1/2) - c^(1/2)
)^5) + (64*a^(3/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^8)/(d^4*((c + d*x)^(1/2) - c^(1/2))^8) - (((a + b*x)^(1
/2) - a^(1/2))^11*((5*a^3*d^3)/4 + (5*b^3*c^3)/4 + (3*a*b^2*c^2*d)/4 + (3*a^2*b*c*d^2)/4))/(b^3*d^4*((c + d*x)
^(1/2) - c^(1/2))^11) + (((a + b*x)^(1/2) - a^(1/2))^9*((85*a^3*d^3)/12 + (85*b^3*c^3)/12 + (17*a*b^2*c^2*d)/4
+ (17*a^2*b*c*d^2)/4))/(b^2*d^5*((c + d*x)^(1/2) - c^(1/2))^9) - (((a + b*x)^(1/2) - a^(1/2))^7*((33*a^3*d^3)
/2 + (33*b^3*c^3)/2 + (327*a*b^2*c^2*d)/2 + (327*a^2*b*c*d^2)/2))/(b*d^6*((c + d*x)^(1/2) - c^(1/2))^7) + (a^(
1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(128*a^2*d^2 + 128*b^2*c^2 + (896*a*b*c*d)/3))/(d^6*((c + d*x)^(1/2
) - c^(1/2))^6) + (64*a^(3/2)*b^2*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^4)/(d^6*((c + d*x)^(1/2) - c^(1/2))^4))/
(((a + b*x)^(1/2) - a^(1/2))^12/((c + d*x)^(1/2) - c^(1/2))^12 + b^6/d^6 - (6*b^5*((a + b*x)^(1/2) - a^(1/2))^
2)/(d^5*((c + d*x)^(1/2) - c^(1/2))^2) + (15*b^4*((a + b*x)^(1/2) - a^(1/2))^4)/(d^4*((c + d*x)^(1/2) - c^(1/2
))^4) - (20*b^3*((a + b*x)^(1/2) - a^(1/2))^6)/(d^3*((c + d*x)^(1/2) - c^(1/2))^6) + (15*b^2*((a + b*x)^(1/2)
- a^(1/2))^8)/(d^2*((c + d*x)^(1/2) - c^(1/2))^8) - (6*b*((a + b*x)^(1/2) - a^(1/2))^10)/(d*((c + d*x)^(1/2) -
c^(1/2))^10)) - (atanh((d^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x)^(1/2) - c^(1/2))))*(a*d + b*
c)*(5*a^2*d^2 + 5*b^2*c^2 - 2*a*b*c*d))/(4*b^(7/2)*d^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {a + b x} \sqrt {c + d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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