3.641 \(\int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=267 \[ \frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{9/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right )}{4 d^4 (b c-a d)}+\frac {(a+b x)^{3/2} \sqrt {c+d x} \left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right )}{6 d^3 (b c-a d)^2}+\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{3 d^2 \sqrt {c+d x} (b c-a d)^2} \]

[Out]

2/3*c^2*(b*x+a)^(5/2)/d^2/(-a*d+b*c)/(d*x+c)^(3/2)+1/4*(3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)*arctanh(d^(1/2)*(b*x+
a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(9/2)/b^(1/2)-4/3*c*(-3*a*d+4*b*c)*(b*x+a)^(5/2)/d^2/(-a*d+b*c)^2/(d*x+c)^(1
/2)+1/6*(3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^3/(-a*d+b*c)^2-1/4*(3*a^2*d^2-30*a*b*c
*d+35*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4/(-a*d+b*c)

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Rubi [A]  time = 0.28, antiderivative size = 267, 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, 78, 50, 63, 217, 206} \[ \frac {(a+b x)^{3/2} \sqrt {c+d x} \left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right )}{6 d^3 (b c-a d)^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right )}{4 d^4 (b c-a d)}+\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{9/2}}+\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{3 d^2 \sqrt {c+d x} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

(2*c^2*(a + b*x)^(5/2))/(3*d^2*(b*c - a*d)*(c + d*x)^(3/2)) - (4*c*(4*b*c - 3*a*d)*(a + b*x)^(5/2))/(3*d^2*(b*
c - a*d)^2*Sqrt[c + d*x]) - ((35*b^2*c^2 - 30*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^4*(b*c -
a*d)) + ((35*b^2*c^2 - 30*a*b*c*d + 3*a^2*d^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(6*d^3*(b*c - a*d)^2) + ((35*b^2
*c^2 - 30*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*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 78

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

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)^{5/2}} \, dx &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {2 \int \frac {(a+b x)^{3/2} \left (\frac {1}{2} c (5 b c-3 a d)-\frac {3}{2} d (b c-a d) x\right )}{(c+d x)^{3/2}} \, dx}{3 d^2 (b c-a d)}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{3 d^2 (b c-a d)^2}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3 (b c-a d)^2}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{4 d^3 (b c-a d)}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4 (b c-a d)}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3 (b c-a d)^2}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 d^4}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4 (b c-a d)}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3 (b c-a d)^2}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\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 )}{4 b d^4}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4 (b c-a d)}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3 (b c-a d)^2}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\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 )}{4 b d^4}\\ &=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4 (b c-a d)}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3 (b c-a d)^2}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 220, normalized size = 0.82 \[ \frac {\frac {3 (c+d x)^2 \left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}}+\frac {\sqrt {d} \left (a^2 d \left (55 c^2+78 c d x+15 d^2 x^2\right )+a b \left (-105 c^3-85 c^2 d x+57 c d^2 x^2+21 d^3 x^3\right )+b^2 x \left (-105 c^3-140 c^2 d x-21 c d^2 x^2+6 d^3 x^3\right )\right )}{\sqrt {a+b x}}}{12 d^{9/2} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

((Sqrt[d]*(a^2*d*(55*c^2 + 78*c*d*x + 15*d^2*x^2) + b^2*x*(-105*c^3 - 140*c^2*d*x - 21*c*d^2*x^2 + 6*d^3*x^3)
+ a*b*(-105*c^3 - 85*c^2*d*x + 57*c*d^2*x^2 + 21*d^3*x^3)))/Sqrt[a + b*x] + (3*(35*b^2*c^2 - 30*a*b*c*d + 3*a^
2*d^2)*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c
- a*d)]))/(12*d^(9/2)*(c + d*x)^(3/2))

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fricas [A]  time = 1.83, size = 594, normalized size = 2.22 \[ \left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (6 \, b^{2} d^{4} x^{3} - 105 \, b^{2} c^{3} d + 55 \, a b c^{2} d^{2} - 3 \, {\left (7 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d^{2} - 39 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b d^{7} x^{2} + 2 \, b c d^{6} x + b c^{2} d^{5}\right )}}, -\frac {3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (6 \, b^{2} d^{4} x^{3} - 105 \, b^{2} c^{3} d + 55 \, a b c^{2} d^{2} - 3 \, {\left (7 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d^{2} - 39 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (b d^{7} x^{2} + 2 \, b c d^{6} x + b c^{2} d^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*(35*b^2*c^4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 30*a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(35*
b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*d^3)*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)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(6*b^2*d^4*x^3 - 105
*b^2*c^3*d + 55*a*b*c^2*d^2 - 3*(7*b^2*c*d^3 - 5*a*b*d^4)*x^2 - 2*(70*b^2*c^2*d^2 - 39*a*b*c*d^3)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(b*d^7*x^2 + 2*b*c*d^6*x + b*c^2*d^5), -1/24*(3*(35*b^2*c^4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2
+ (35*b^2*c^2*d^2 - 30*a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(35*b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*d^3)*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*(6*b^2*d^4*x^3 - 105*b^2*c^3*d + 55*a*b*c^2*d^2 - 3*(7*b^2*c*d^3 - 5*a*b*d^4)*x^2 - 2*(70*
b^2*c^2*d^2 - 39*a*b*c*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^7*x^2 + 2*b*c*d^6*x + b*c^2*d^5)]

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giac [A]  time = 2.07, size = 393, normalized size = 1.47 \[ \frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{6} - a b^{5} d^{7}\right )} {\left (b x + a\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}} - \frac {7 \, b^{7} c^{2} d^{5} - 6 \, a b^{6} c d^{6} - a^{2} b^{5} d^{7}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} - \frac {4 \, {\left (35 \, b^{8} c^{3} d^{4} - 65 \, a b^{7} c^{2} d^{5} + 33 \, a^{2} b^{6} c d^{6} - 3 \, a^{3} b^{5} d^{7}\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (35 \, b^{9} c^{4} d^{3} - 100 \, a b^{8} c^{3} d^{4} + 98 \, a^{2} b^{7} c^{2} d^{5} - 36 \, a^{3} b^{6} c d^{6} + 3 \, a^{4} b^{5} d^{7}\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {{\left (35 \, b^{3} c^{2} - 30 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{4} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/12*((3*(b*x + a)*(2*(b^6*c*d^6 - a*b^5*d^7)*(b*x + a)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)) - (7*b^7*c^2*d^5
- 6*a*b^6*c*d^6 - a^2*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b))) - 4*(35*b^8*c^3*d^4 - 65*a*b^7*c^2*d^5
+ 33*a^2*b^6*c*d^6 - 3*a^3*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)))*(b*x + a) - 3*(35*b^9*c^4*d^3 - 100
*a*b^8*c^3*d^4 + 98*a^2*b^7*c^2*d^5 - 36*a^3*b^6*c*d^6 + 3*a^4*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)))
*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(35*b^3*c^2 - 30*a*b^2*c*d + 3*a^2*b*d^2)*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 = 676, normalized size = 2.53 \[ \frac {\sqrt {b x +a}\, \left (9 a^{2} d^{4} x^{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 )-90 a b c \,d^{3} x^{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 )+105 b^{2} c^{2} d^{2} x^{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 )+18 a^{2} 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 )-180 a b \,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 )+210 b^{2} 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 )+9 a^{2} 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 )-90 a b \,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^{2} 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 )+12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,d^{3} x^{3}+30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,d^{3} x^{2}-42 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b c \,d^{2} x^{2}+156 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a c \,d^{2} x -280 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{2} d x +110 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,c^{2} d -210 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{3}\right )}{24 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x)

[Out]

1/24*(b*x+a)^(1/2)*(9*a^2*d^4*x^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))-
90*a*b*c*d^3*x^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))+105*b^2*c^2*d^2*x
^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))+12*((b*x+a)*(d*x+c))^(1/2)*(b*d
)^(1/2)*x^3*b*d^3+18*a^2*c*d^3*x*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))-1
80*a*b*c^2*d^2*x*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))+210*b^2*c^3*d*x*l
n(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))+30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1
/2)*a*d^3*x^2-42*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c*d^2*x^2+9*a^2*c^2*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))-90*a*b*c^3*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))+105*b^2*c^4*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)
)+156*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c*d^2*x-280*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c^2*d*x+110*((b*
x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c^2*d-210*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c^3)/(b*d)^(1/2)/((b*x+a)*(d
*x+c))^(1/2)/d^4/(d*x+c)^(3/2)

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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^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),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 \[ \int \frac {x^2\,{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2),x)

[Out]

int((x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**(3/2)/(d*x+c)**(5/2),x)

[Out]

Timed out

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