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

Optimal. Leaf size=191 \[ \frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+2 a b c d+5 b^2 c^2\right )}{8 b^2 d^3}-\frac {(b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} d^{7/2}}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (3 a d+5 b c)}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d} \]

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Rubi [A]  time = 0.16, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {90, 80, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+2 a b c d+5 b^2 c^2\right )}{8 b^2 d^3}-\frac {(b c-a d) \left (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^{5/2} d^{7/2}}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (3 a d+5 b c)}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^2*d^3) - ((5*b*c + 3*a*d)*(a + b*x)^(3/2)
*Sqrt[c + d*x])/(12*b^2*d^2) + (x*(a + b*x)^(3/2)*Sqrt[c + d*x])/(3*b*d) - ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d
+ a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(5/2)*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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 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^2 \sqrt {a+b x}}{\sqrt {c+d x}} \, dx &=\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d}+\frac {\int \frac {\sqrt {a+b x} \left (-a c-\frac {1}{2} (5 b c+3 a d) x\right )}{\sqrt {c+d x}} \, dx}{3 b d}\\ &=-\frac {(5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d}+\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 b^2 d^2}\\ &=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^2 d^3}-\frac {(5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b^2 d^3}\\ &=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^2 d^3}-\frac {(5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+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^3 d^3}\\ &=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^2 d^3}-\frac {(5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+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^3 d^3}\\ &=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^2 d^3}-\frac {(5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b^2 d^2}+\frac {x (a+b x)^{3/2} \sqrt {c+d x}}{3 b d}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 173, normalized size = 0.91 \begin {gather*} \frac {-b \sqrt {d} \sqrt {a+b x} (c+d x) \left (3 a^2 d^2-2 a b d (d x-2 c)+b^2 \left (-15 c^2+10 c d x-8 d^2 x^2\right )\right )-3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) (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 )}{24 b^3 d^{7/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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IntegrateAlgebraic [A]  time = 0.67, size = 228, normalized size = 1.19 \begin {gather*} \frac {\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (-3 a^2 d^2 \sqrt {c+d x}+2 a b d (c+d x)^{3/2}-6 a b c d \sqrt {c+d x}+33 b^2 c^2 \sqrt {c+d x}+8 b^2 (c+d x)^{5/2}-26 b^2 c (c+d x)^{3/2}\right )}{24 b^2 d^3}+\frac {\sqrt {\frac {b}{d}} \left (-a^3 d^3-a^2 b c d^2-3 a b^2 c^2 d+5 b^3 c^3\right ) \log \left (\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{8 b^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - (b*c)/d + (b*(c + d*x))/d]*(33*b^2*c^2*Sqrt[c + d*x] - 6*a*b*c*d*Sqrt[c + d*x] - 3*a^2*d^2*Sqrt[c +
d*x] - 26*b^2*c*(c + d*x)^(3/2) + 2*a*b*d*(c + d*x)^(3/2) + 8*b^2*(c + d*x)^(5/2)))/(24*b^2*d^3) + (Sqrt[b/d]*
(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*Log[-(Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[a - (b*c)/d + (b*(c
+ d*x))/d]])/(8*b^3*d^3)

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fricas [A]  time = 1.36, size = 414, normalized size = 2.17 \begin {gather*} \left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 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 - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{3} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 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 - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{3} d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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 - 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 - 4*a*b^2*c*d^2 - 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d
*x + c))/(b^3*d^4), 1/48*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - 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 - 4*a*b^2*c*d^2 - 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*
x + c))/(b^3*d^4)]

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giac [A]  time = 1.83, size = 214, normalized size = 1.12 \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^{3} d} - \frac {5 \, b^{7} c d^{3} + 7 \, a b^{6} d^{4}}{b^{9} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{8} c^{2} d^{2} + 2 \, a b^{7} c d^{3} + a^{2} b^{6} d^{4}\right )}}{b^{9} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 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^{2} d^{3}}\right )} b}{24 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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^3*d) - (5*b^7*c*d^3 + 7*a
*b^6*d^4)/(b^9*d^5)) + 3*(5*b^8*c^2*d^2 + 2*a*b^7*c*d^3 + a^2*b^6*d^4)/(b^9*d^5)) + 3*(5*b^3*c^3 - 3*a*b^2*c^2
*d - 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^2*d^3))*b/abs(b)

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maple [B]  time = 0.02, size = 395, normalized size = 2.07 \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 )+3 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}+4 \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}-8 \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^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*x^2*b^2*d^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+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^3*d^3+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^2*b*c*d^2+9*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*d-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))*b^3*c^3+4*
(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b*d^2-20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*b^2*c*d-6*(b*d)^(1/2)*(
(b*x+a)*(d*x+c))^(1/2)*a^2*d^2-8*(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)/d^3/((b*x+a)*(d*x+c))^(1/2)/b^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^2*(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 = 50.04, size = 929, normalized size = 4.86 \begin {gather*} \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 (a^2\,d^2+2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{4\,b^{5/2}\,d^{7/2}}-\frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {a^3\,b^3\,d^3}{4}+\frac {a^2\,b^4\,c\,d^2}{4}+\frac {3\,a\,b^5\,c^2\,d}{4}-\frac {5\,b^6\,c^3}{4}\right )}{d^9\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {17\,a^3\,b^2\,d^3}{12}+\frac {91\,a^2\,b^3\,c\,d^2}{4}+\frac {17\,a\,b^4\,c^2\,d}{4}-\frac {85\,b^5\,c^3}{12}\right )}{d^8\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {19\,a^3\,d^3}{2}+\frac {275\,a^2\,b\,c\,d^2}{2}+\frac {313\,a\,b^2\,c^2\,d}{2}+\frac {33\,b^3\,c^3}{2}\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {19\,a^3\,b\,d^3}{2}+\frac {275\,a^2\,b^2\,c\,d^2}{2}+\frac {313\,a\,b^3\,c^2\,d}{2}+\frac {33\,b^4\,c^3}{2}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (\frac {a^3\,d^3}{4}+\frac {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^2\,d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (\frac {17\,a^3\,d^3}{12}+\frac {91\,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\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}+\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,d\,a^2+96\,b\,c\,a\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,d\,a^2\,b^2+96\,c\,a\,b^3\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (64\,a^2\,b\,d^2+\frac {704\,a\,b^2\,c\,d}{3}+128\,b^3\,c^2\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}}{\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)*(a^2*d^2 + 5*b
^2*c^2 + 2*a*b*c*d))/(4*b^(5/2)*d^(7/2)) - ((((a + b*x)^(1/2) - a^(1/2))*((a^3*b^3*d^3)/4 - (5*b^6*c^3)/4 + (a
^2*b^4*c*d^2)/4 + (3*a*b^5*c^2*d)/4))/(d^9*((c + d*x)^(1/2) - c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^3*((17*
a^3*b^2*d^3)/12 - (85*b^5*c^3)/12 + (91*a^2*b^3*c*d^2)/4 + (17*a*b^4*c^2*d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2
))^3) - (((a + b*x)^(1/2) - a^(1/2))^7*((19*a^3*d^3)/2 + (33*b^3*c^3)/2 + (313*a*b^2*c^2*d)/2 + (275*a^2*b*c*d
^2)/2))/(d^6*((c + d*x)^(1/2) - c^(1/2))^7) - (((a + b*x)^(1/2) - a^(1/2))^5*((33*b^4*c^3)/2 + (19*a^3*b*d^3)/
2 + (275*a^2*b^2*c*d^2)/2 + (313*a*b^3*c^2*d)/2))/(d^7*((c + d*x)^(1/2) - c^(1/2))^5) + (((a + b*x)^(1/2) - a^
(1/2))^11*((a^3*d^3)/4 - (5*b^3*c^3)/4 + (3*a*b^2*c^2*d)/4 + (a^2*b*c*d^2)/4))/(b^2*d^4*((c + d*x)^(1/2) - c^(
1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^9*((17*a^3*d^3)/12 - (85*b^3*c^3)/12 + (17*a*b^2*c^2*d)/4 + (91*a^2*b
*c*d^2)/4))/(b*d^5*((c + d*x)^(1/2) - c^(1/2))^9) + (a^(1/2)*c^(1/2)*(32*a^2*d + 96*a*b*c)*((a + b*x)^(1/2) -
a^(1/2))^8)/(d^4*((c + d*x)^(1/2) - c^(1/2))^8) + (a^(1/2)*c^(1/2)*(32*a^2*b^2*d + 96*a*b^3*c)*((a + b*x)^(1/2
) - a^(1/2))^4)/(d^6*((c + d*x)^(1/2) - c^(1/2))^4) + (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(128*b^3*
c^2 + 64*a^2*b*d^2 + (704*a*b^2*c*d)/3))/(d^6*((c + d*x)^(1/2) - c^(1/2))^6))/(((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))

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

[Out]

Timed out

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