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

Optimal. Leaf size=248 \[ \frac {\sqrt {a+b x} \sqrt {c+d x} \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right )}{8 b d^4}-\frac {(b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{9/2}}+\frac {(a+b x)^{3/2} \sqrt {c+d x} \left (\frac {a^2 d}{b}+10 a c-\frac {35 b c^2}{d}\right )}{12 d^2 (b c-a d)}+\frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2} \]

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Rubi [A]  time = 0.25, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {89, 80, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right )}{8 b d^4}-\frac {(b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{9/2}}+\frac {(a+b x)^{3/2} \sqrt {c+d x} \left (\frac {a^2 d}{b}+10 a c-\frac {35 b c^2}{d}\right )}{12 d^2 (b c-a d)}+\frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(a + b*x)^(5/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) + ((35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*Sqrt[a + b*x]*S
qrt[c + d*x])/(8*b*d^4) + ((10*a*c - (35*b*c^2)/d + (a^2*d)/b)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(12*d^2*(b*c - a
*d)) + ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*b*d^2) - ((b*c - a*d)*(35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*ArcTanh[(S
qrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(3/2)*d^(9/2))

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {2 \int \frac {(a+b x)^{3/2} \left (\frac {1}{2} c (5 b c-a d)-\frac {1}{2} d (b c-a d) x\right )}{\sqrt {c+d x}} \, dx}{d^2 (b c-a d)}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}-\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{6 b d^2 (b c-a d)}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^3 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}+\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 b d^3}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^4}-\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^3 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}-\frac {\left ((b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b d^4}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^4}-\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^3 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}-\frac {\left ((b c-a d) \left (35 b^2 c^2-10 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^2 d^4}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^4}-\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^3 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}-\frac {\left ((b c-a d) \left (35 b^2 c^2-10 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^2 d^4}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^4}-\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^3 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}-\frac {(b c-a d) \left (35 b^2 c^2-10 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^{3/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 233, normalized size = 0.94 \begin {gather*} \frac {\frac {b \sqrt {d} \left (3 a^3 d^2 (c+d x)+a^2 b d \left (-100 c^2-35 c d x+17 d^2 x^2\right )+a b^2 \left (105 c^3-65 c^2 d x-52 c d^2 x^2+22 d^3 x^3\right )+b^3 x \left (105 c^3+35 c^2 d x-14 c d^2 x^2+8 d^3 x^3\right )\right )}{\sqrt {a+b x}}-3 (b c-a d)^{3/2} \left (-a^2 d^2-10 a b c d+35 b^2 c^2\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^2 d^{9/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[d]*(3*a^3*d^2*(c + d*x) + a^2*b*d*(-100*c^2 - 35*c*d*x + 17*d^2*x^2) + b^3*x*(105*c^3 + 35*c^2*d*x -
14*c*d^2*x^2 + 8*d^3*x^3) + a*b^2*(105*c^3 - 65*c^2*d*x - 52*c*d^2*x^2 + 22*d^3*x^3)))/Sqrt[a + b*x] - 3*(b*c
- a*d)^(3/2)*(35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*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^(9/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 1.62, size = 234, normalized size = 0.94 \begin {gather*} \frac {\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (3 a^2 d^2 (c+d x)-48 a b c^2 d-66 a b c d (c+d x)+14 a b d (c+d x)^2+48 b^2 c^3+87 b^2 c^2 (c+d x)-38 b^2 c (c+d x)^2+8 b^2 (c+d x)^3\right )}{24 b d^4 \sqrt {c+d x}}+\frac {\sqrt {\frac {b}{d}} \left (a^3 d^3+9 a^2 b c d^2-45 a b^2 c^2 d+35 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^2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - (b*c)/d + (b*(c + d*x))/d]*(48*b^2*c^3 - 48*a*b*c^2*d + 87*b^2*c^2*(c + d*x) - 66*a*b*c*d*(c + d*x)
+ 3*a^2*d^2*(c + d*x) - 38*b^2*c*(c + d*x)^2 + 14*a*b*d*(c + d*x)^2 + 8*b^2*(c + d*x)^3))/(24*b*d^4*Sqrt[c + d
*x]) + (Sqrt[b/d]*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*Log[-(Sqrt[b/d]*Sqrt[c + d*x]) + Sqr
t[a - (b*c)/d + (b*(c + d*x))/d]])/(8*b^2*d^4)

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fricas [A]  time = 1.86, size = 600, normalized size = 2.42 \begin {gather*} \left [\frac {3 \, {\left (35 \, b^{3} c^{4} - 45 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3} + {\left (35 \, b^{3} c^{3} d - 45 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} + a^{3} d^{4}\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 (8 \, b^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 100 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 14 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 38 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{2} d^{6} x + b^{2} c d^{5}\right )}}, \frac {3 \, {\left (35 \, b^{3} c^{4} - 45 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3} + {\left (35 \, b^{3} c^{3} d - 45 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} + a^{3} d^{4}\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 (8 \, b^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 100 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 14 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 38 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{2} d^{6} x + b^{2} c d^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(35*b^3*c^4 - 45*a*b^2*c^3*d + 9*a^2*b*c^2*d^2 + a^3*c*d^3 + (35*b^3*c^3*d - 45*a*b^2*c^2*d^2 + 9*a^2
*b*c*d^3 + a^3*d^4)*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)*s
qrt(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^4*x^3 + 105*b^3*c^3*d - 100*a*b^2
*c^2*d^2 + 3*a^2*b*c*d^3 - 14*(b^3*c*d^3 - a*b^2*d^4)*x^2 + (35*b^3*c^2*d^2 - 38*a*b^2*c*d^3 + 3*a^2*b*d^4)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^6*x + b^2*c*d^5), 1/48*(3*(35*b^3*c^4 - 45*a*b^2*c^3*d + 9*a^2*b*c^2*d^2
+ a^3*c*d^3 + (35*b^3*c^3*d - 45*a*b^2*c^2*d^2 + 9*a^2*b*c*d^3 + a^3*d^4)*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*(8*b^3*
d^4*x^3 + 105*b^3*c^3*d - 100*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 14*(b^3*c*d^3 - a*b^2*d^4)*x^2 + (35*b^3*c^2*d^2
- 38*a*b^2*c*d^3 + 3*a^2*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^6*x + b^2*c*d^5)]

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giac [A]  time = 1.62, size = 282, normalized size = 1.14 \begin {gather*} \frac {{\left ({\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{d {\left | b \right |}} - \frac {7 \, b^{3} c d^{5} + 5 \, a b^{2} d^{6}}{b^{2} d^{7} {\left | b \right |}}\right )} + \frac {35 \, b^{4} c^{2} d^{4} - 10 \, a b^{3} c d^{5} - a^{2} b^{2} d^{6}}{b^{2} d^{7} {\left | b \right |}}\right )} + \frac {3 \, {\left (35 \, b^{5} c^{3} d^{3} - 45 \, a b^{4} c^{2} d^{4} + 9 \, a^{2} b^{3} c d^{5} + a^{3} b^{2} d^{6}\right )}}{b^{2} d^{7} {\left | b \right |}}\right )} \sqrt {b x + a}}{24 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, 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 )}{8 \, \sqrt {b d} d^{4} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(d*abs(b)) - (7*b^3*c*d^5 + 5*a*b^2*d^6)/(b^2*d^7*abs(b))) + (35*b^4
*c^2*d^4 - 10*a*b^3*c*d^5 - a^2*b^2*d^6)/(b^2*d^7*abs(b))) + 3*(35*b^5*c^3*d^3 - 45*a*b^4*c^2*d^4 + 9*a^2*b^3*
c*d^5 + a^3*b^2*d^6)/(b^2*d^7*abs(b)))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 1/8*(35*b^3*c^3 - 4
5*a*b^2*c^2*d + 9*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)*d^4*abs(b))

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maple [B]  time = 0.03, size = 692, normalized size = 2.79 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (3 a^{3} d^{4} 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 )+27 a^{2} b c \,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 )-135 a \,b^{2} c^{2} 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 )+105 b^{3} c^{3} 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^{3} c \,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 )+27 a^{2} b \,c^{2} 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 )-135 a \,b^{2} c^{3} 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 )+105 b^{3} c^{4} \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 {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} d^{3} x^{3}-28 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,d^{3} x^{2}+28 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c \,d^{2} x^{2}-6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} d^{3} x +76 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b c \,d^{2} x -70 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{2} d x -6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} c \,d^{2}+200 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,c^{2} d -210 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{3}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {d x +c}\, b \,d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(b*x+a)^(1/2)*(-16*x^3*b^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(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))*x*a^3*d^4+27*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))*x*a^2*b*c*d^3-135*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))*x*a*b^2*c^2*d^2+105*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))*x*b^3*c^3*
d-28*x^2*a*b*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+28*x^2*b^2*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(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*c*d^3+27*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^2*d^2-135*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^3*d+105*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^4-6*x*a^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+76*x*a*b*c*d^2*((b*x+a)*(d*x+c))^(1
/2)*(b*d)^(1/2)-70*x*b^2*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-6*a^2*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(
1/2)+200*a*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-210*b^2*c^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+
a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(d*x+c)^(1/2)/b/d^4

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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)^(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 = -0.00 \begin {gather*} \int \frac {x^2\,{\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^2*(a + b*x)^(3/2))/(c + d*x)^(3/2),x)

[Out]

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

[Out]

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

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