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

Optimal. Leaf size=165 \[ \frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac{4 c (a+b x)^{3/2}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{a+b x} \sqrt{c+d x} (5 b c-a d)}{d^3 (b c-a d)}-\frac{(5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{b} d^{7/2}} \]

[Out]

(2*c^2*(a + b*x)^(3/2))/(3*d^2*(b*c - a*d)*(c + d*x)^(3/2)) - (4*c*(a + b*x)^(3/2))/(d^2*(b*c - a*d)*Sqrt[c +
d*x]) + ((5*b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(d^3*(b*c - a*d)) - ((5*b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a
+ b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(7/2))

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Rubi [A]  time = 0.161388, antiderivative size = 165, normalized size of antiderivative = 1., 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 = {89, 78, 50, 63, 217, 206} \[ \frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac{4 c (a+b x)^{3/2}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{a+b x} \sqrt{c+d x} (5 b c-a d)}{d^3 (b c-a d)}-\frac{(5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{b} d^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(a + b*x)^(3/2))/(3*d^2*(b*c - a*d)*(c + d*x)^(3/2)) - (4*c*(a + b*x)^(3/2))/(d^2*(b*c - a*d)*Sqrt[c +
d*x]) + ((5*b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(d^3*(b*c - a*d)) - ((5*b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a
+ b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(7/2))

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 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 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 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]

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])

Rubi steps

\begin{align*} \int \frac{x^2 \sqrt{a+b x}}{(c+d x)^{5/2}} \, dx &=\frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac{2 \int \frac{\sqrt{a+b x} \left (\frac{3}{2} c (b c-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)^{3/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac{4 c (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{d^2 (b c-a d)}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac{4 c (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{d^3 (b c-a d)}-\frac{(5 b c-a d) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 d^3}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac{4 c (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{d^3 (b c-a d)}-\frac{(5 b c-a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{b d^3}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac{4 c (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{d^3 (b c-a d)}-\frac{(5 b c-a d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{b d^3}\\ &=\frac{2 c^2 (a+b x)^{3/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac{4 c (a+b x)^{3/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{d^3 (b c-a d)}-\frac{(5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{b} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.669227, size = 203, normalized size = 1.23 \[ \frac{\frac{\sqrt{d} \left (a^2 (-d) \left (13 c^2+18 c d x+3 d^2 x^2\right )+a b \left (7 c^2 d x+15 c^3-15 c d^2 x^2-3 d^3 x^3\right )+b^2 c x \left (15 c^2+20 c d x+3 d^2 x^2\right )\right )}{\sqrt{a+b x} (b c-a d)}-\frac{3 (b c-a d)^{3/2} (5 b c-a d) \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b^2}}{3 d^{7/2} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d]*(-(a^2*d*(13*c^2 + 18*c*d*x + 3*d^2*x^2)) + b^2*c*x*(15*c^2 + 20*c*d*x + 3*d^2*x^2) + a*b*(15*c^3 +
7*c^2*d*x - 15*c*d^2*x^2 - 3*d^3*x^3)))/((b*c - a*d)*Sqrt[a + b*x]) - (3*(b*c - a*d)^(3/2)*(5*b*c - a*d)*((b*(
c + d*x))/(b*c - a*d))^(3/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/b^2)/(3*d^(7/2)*(c + d*x)^(3/2)
)

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Maple [B]  time = 0.019, size = 659, normalized size = 4. \begin{align*}{\frac{1}{ \left ( 6\,ad-6\,bc \right ){d}^{3}} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{a}^{2}{d}^{4}-18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}abc{d}^{3}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{2}{c}^{2}{d}^{2}+6\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) x{a}^{2}c{d}^{3}-36\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) xab{c}^{2}{d}^{2}+30\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) x{b}^{2}{c}^{3}d+6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}{x}^{2}a{d}^{3}-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}{x}^{2}bc{d}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}{c}^{2}{d}^{2}-18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) ab{c}^{3}d+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{2}{c}^{4}+36\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}xac{d}^{2}-40\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}xb{c}^{2}d+26\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}a{c}^{2}d-30\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}b{c}^{3} \right ) \sqrt{bx+a}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^(1/2)/(d*x+c)^(5/2),x)

[Out]

1/6*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^2*d^4-18*ln(1/2*(2*b*
d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a*b*c*d^3+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d
*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*b^2*c^2*d^2+6*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*
d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a^2*c*d^3-36*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(
b*d)^(1/2))*x*a*b*c^2*d^2+30*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^2
*c^3*d+6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^2*a*d^3-6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^2*b*c*d^2+3*ln(
1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c^2*d^2-18*ln(1/2*(2*b*d*x+2*((b*
x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c^3*d+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(
b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^4+36*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*a*c*d^2-40*((b*x+a)*(d*x+c))
^(1/2)*(b*d)^(1/2)*x*b*c^2*d+26*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c^2*d-30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(
1/2)*b*c^3)*(b*x+a)^(1/2)/(a*d-b*c)/(b*d)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/d^3/(d*x+c)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.4912, size = 1368, normalized size = 8.29 \begin{align*} \left [-\frac{3 \,{\left (5 \, b^{2} c^{4} - 6 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (5 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \,{\left (5 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 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 (15 \, b^{2} c^{3} d - 13 \, a b c^{2} d^{2} + 3 \,{\left (b^{2} c d^{3} - a b d^{4}\right )} x^{2} + 2 \,{\left (10 \, b^{2} c^{2} d^{2} - 9 \, a b c d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{12 \,{\left (b^{2} c^{3} d^{4} - a b c^{2} d^{5} +{\left (b^{2} c d^{6} - a b d^{7}\right )} x^{2} + 2 \,{\left (b^{2} c^{2} d^{5} - a b c d^{6}\right )} x\right )}}, \frac{3 \,{\left (5 \, b^{2} c^{4} - 6 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (5 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \,{\left (5 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 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 (15 \, b^{2} c^{3} d - 13 \, a b c^{2} d^{2} + 3 \,{\left (b^{2} c d^{3} - a b d^{4}\right )} x^{2} + 2 \,{\left (10 \, b^{2} c^{2} d^{2} - 9 \, a b c d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (b^{2} c^{3} d^{4} - a b c^{2} d^{5} +{\left (b^{2} c d^{6} - a b d^{7}\right )} x^{2} + 2 \,{\left (b^{2} c^{2} d^{5} - a b c d^{6}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*b^2*c^4 - 6*a*b*c^3*d + a^2*c^2*d^2 + (5*b^2*c^2*d^2 - 6*a*b*c*d^3 + a^2*d^4)*x^2 + 2*(5*b^2*c^3*
d - 6*a*b*c^2*d^2 + 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*(15*b^2*c^3*d - 13*a*b*c^2*d^2
+ 3*(b^2*c*d^3 - a*b*d^4)*x^2 + 2*(10*b^2*c^2*d^2 - 9*a*b*c*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*c^3*d^4
- a*b*c^2*d^5 + (b^2*c*d^6 - a*b*d^7)*x^2 + 2*(b^2*c^2*d^5 - a*b*c*d^6)*x), 1/6*(3*(5*b^2*c^4 - 6*a*b*c^3*d +
a^2*c^2*d^2 + (5*b^2*c^2*d^2 - 6*a*b*c*d^3 + a^2*d^4)*x^2 + 2*(5*b^2*c^3*d - 6*a*b*c^2*d^2 + a^2*c*d^3)*x)*sqr
t(-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*(15*b^2*c^3*d - 13*a*b*c^2*d^2 + 3*(b^2*c*d^3 - a*b*d^4)*x^2 + 2*(10*b^2*c^2*d^2 - 9*a*
b*c*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*c^3*d^4 - a*b*c^2*d^5 + (b^2*c*d^6 - a*b*d^7)*x^2 + 2*(b^2*c^2*d
^5 - a*b*c*d^6)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**(1/2)/(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.41689, size = 382, normalized size = 2.32 \begin{align*} \frac{{\left ({\left (b x + a\right )}{\left (\frac{3 \,{\left (b^{6} c d^{4} - a b^{5} d^{5}\right )}{\left (b x + a\right )}}{b^{4} c d^{5}{\left | b \right |} - a b^{3} d^{6}{\left | b \right |}} + \frac{2 \,{\left (10 \, b^{7} c^{2} d^{3} - 12 \, a b^{6} c d^{4} + 3 \, a^{2} b^{5} d^{5}\right )}}{b^{4} c d^{5}{\left | b \right |} - a b^{3} d^{6}{\left | b \right |}}\right )} + \frac{3 \,{\left (5 \, b^{8} c^{3} d^{2} - 11 \, a b^{7} c^{2} d^{3} + 7 \, a^{2} b^{6} c d^{4} - a^{3} b^{5} d^{5}\right )}}{b^{4} c d^{5}{\left | b \right |} - a b^{3} d^{6}{\left | b \right |}}\right )} \sqrt{b x + a}}{3 \,{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} + \frac{{\left (5 \, b^{2} c - a b d\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} d^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/3*((b*x + a)*(3*(b^6*c*d^4 - a*b^5*d^5)*(b*x + a)/(b^4*c*d^5*abs(b) - a*b^3*d^6*abs(b)) + 2*(10*b^7*c^2*d^3
- 12*a*b^6*c*d^4 + 3*a^2*b^5*d^5)/(b^4*c*d^5*abs(b) - a*b^3*d^6*abs(b))) + 3*(5*b^8*c^3*d^2 - 11*a*b^7*c^2*d^3
+ 7*a^2*b^6*c*d^4 - a^3*b^5*d^5)/(b^4*c*d^5*abs(b) - a*b^3*d^6*abs(b)))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d
- a*b*d)^(3/2) + (5*b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(s
qrt(b*d)*d^3*abs(b))