3.7.99 \(\int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx\) [699]

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

[Out]

1/8*(-a*d+b*c)*(5*a^2*d^2+2*a*b*c*d+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^(5
/2)-1/12*(5*a*d+3*b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b^2/d^2+1/3*x*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b/d+1/8*(5*a^2*d^
2+2*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^2

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Rubi [A]
time = 0.11, 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 = {92, 81, 52, 65, 223, 212} \begin {gather*} \frac {(b c-a d) \left (5 a^2 d^2+2 a b c d+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^{5/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+2 a b c d+b^2 c^2\right )}{8 b^3 d^2}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{12 b^2 d^2}+\frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^3*d^2) - ((3*b*c + 5*a*d)*Sqrt[a + b*x]*(
c + d*x)^(3/2))/(12*b^2*d^2) + (x*Sqrt[a + b*x]*(c + d*x)^(3/2))/(3*b*d) + ((b*c - a*d)*(b^2*c^2 + 2*a*b*c*d +
5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(7/2)*d^(5/2))

Rule 52

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 65

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 81

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 92

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 212

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

Rule 223

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

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Mathematica [A]
time = 0.47, size = 153, normalized size = 0.80 \begin {gather*} \frac {b \sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-2 a b d (2 c+5 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )-3 \sqrt {\frac {b}{d}} \left (b^3 c^3+a b^2 c^2 d+3 a^2 b c d^2-5 a^3 d^3\right ) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{24 b^4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 - 2*a*b*d*(2*c + 5*d*x) + b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)) - 3*S
qrt[b/d]*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*Log[Sqrt[a + b*x] - Sqrt[b/d]*Sqrt[c + d*x]])/(24
*b^4*d^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(159)=318\).
time = 0.07, size = 395, normalized size = 2.07

method	result	size
default	\(-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (-16 b^{2} d^{2} x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}-9 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+20 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a b \,d^{2} x -4 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c d x -30 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} d^{2}+8 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a b c d +6 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} d^{2} \sqrt {b d}}\)	\(395\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-16*b^2*d^2*x^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+15*ln(1/2*(2*b*d*x+2*((
d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*d^3-9*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b
*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c*d^2-3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(
b*d)^(1/2))*a*b^2*c^2*d-3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^3+
20*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b*d^2*x-4*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^2*c*d*x-30*(b*d)^(1/2
)*((d*x+c)*(b*x+a))^(1/2)*a^2*d^2+8*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b*c*d+6*(b*d)^(1/2)*((d*x+c)*(b*x+a)
)^(1/2)*b^2*c^2)/((d*x+c)*(b*x+a))^(1/2)/b^3/d^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*(d*x+c)^(1/2)/(b*x+a)^(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 detail

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Fricas [A]
time = 0.72, size = 408, normalized size = 2.14 \begin {gather*} \left [-\frac {3 \, {\left (b^{3} c^{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} - 3 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{3}}, -\frac {3 \, {\left (b^{3} c^{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} - 3 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
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*(d*x+c)**(1/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 1.63, size = 207, normalized size = 1.08 \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}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{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 d^{2}}\right )} {\left | b \right |}}{24 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x+c)^(1/2)/(b*x+a)^(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 + (b^6*c*d^3 - 13*a*b^5*
d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^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*
d^2))*abs(b)/b^3

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Mupad [B]
time = 50.20, size = 924, normalized size = 4.84 \begin {gather*} -\frac {\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 {a\,b^4\,c^2\,d}{4}+\frac {b^5\,c^3}{4}\right )}{d^8\,\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 {313\,a^2\,b\,c\,d^2}{2}+\frac {275\,a\,b^2\,c^2\,d}{2}+\frac {19\,b^3\,c^3}{2}\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}-\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 {91\,a\,b^3\,c^2\,d}{4}+\frac {17\,b^4\,c^3}{12}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}+\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 {a\,b^2\,c^2\,d}{4}+\frac {b^3\,c^3}{4}\right )}{b^3\,d^3\,{\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 {91\,a\,b^2\,c^2\,d}{4}+\frac {17\,b^3\,c^3}{12}\right )}{b^2\,d^4\,{\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 {313\,a^2\,b\,c\,d^2}{2}+\frac {275\,a\,b^2\,c^2\,d}{2}+\frac {19\,b^3\,c^3}{2}\right )}{b\,d^5\,{\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 {704\,a\,b\,c\,d}{3}+64\,b^2\,c^2\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,b\,c^2+96\,a\,d\,c\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\,b^3\,c^2+96\,a\,d\,b^2\,c\right )\,{\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+b^2\,c^2\right )}{4\,b^{7/2}\,d^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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