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

Optimal. Leaf size=319 \[ \frac{\sqrt{a+b x} (c+d x)^{5/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{3 b^3 (b c-a d)^2}+\frac{5 \sqrt{a+b x} (c+d x)^{3/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{12 b^4 (b c-a d)}+\frac{5 \sqrt{a+b x} \sqrt{c+d x} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{8 b^5}+\frac{5 (b c-a d) \left (21 a^2 d^2-14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{11/2} \sqrt{d}}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (a+b x)^{3/2} (b c-a d)}+\frac{4 a (c+d x)^{7/2} (3 b c-5 a d)}{3 b^2 \sqrt{a+b x} (b c-a d)^2} \]

[Out]

(5*(b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^5) + (5*(b^2*c^2 - 14*a*b*c*d + 21*a^
2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b^4*(b*c - a*d)) + ((b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[a + b*x
]*(c + d*x)^(5/2))/(3*b^3*(b*c - a*d)^2) - (2*a^2*(c + d*x)^(7/2))/(3*b^2*(b*c - a*d)*(a + b*x)^(3/2)) + (4*a*
(3*b*c - 5*a*d)*(c + d*x)^(7/2))/(3*b^2*(b*c - a*d)^2*Sqrt[a + b*x]) + (5*(b*c - a*d)*(b^2*c^2 - 14*a*b*c*d +
21*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(11/2)*Sqrt[d])

________________________________________________________________________________________

Rubi [A]  time = 0.326724, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{\sqrt{a+b x} (c+d x)^{5/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{3 b^3 (b c-a d)^2}+\frac{5 \sqrt{a+b x} (c+d x)^{3/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{12 b^4 (b c-a d)}+\frac{5 \sqrt{a+b x} \sqrt{c+d x} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{8 b^5}+\frac{5 (b c-a d) \left (21 a^2 d^2-14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{11/2} \sqrt{d}}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (a+b x)^{3/2} (b c-a d)}+\frac{4 a (c+d x)^{7/2} (3 b c-5 a d)}{3 b^2 \sqrt{a+b x} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^5) + (5*(b^2*c^2 - 14*a*b*c*d + 21*a^
2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b^4*(b*c - a*d)) + ((b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[a + b*x
]*(c + d*x)^(5/2))/(3*b^3*(b*c - a*d)^2) - (2*a^2*(c + d*x)^(7/2))/(3*b^2*(b*c - a*d)*(a + b*x)^(3/2)) + (4*a*
(3*b*c - 5*a*d)*(c + d*x)^(7/2))/(3*b^2*(b*c - a*d)^2*Sqrt[a + b*x]) + (5*(b*c - a*d)*(b^2*c^2 - 14*a*b*c*d +
21*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(11/2)*Sqrt[d])

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 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx &=-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{2 \int \frac{(c+d x)^{5/2} \left (-\frac{1}{2} a (3 b c-7 a d)+\frac{3}{2} b (b c-a d) x\right )}{(a+b x)^{3/2}} \, dx}{3 b^2 (b c-a d)}\\ &=-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \int \frac{(c+d x)^{5/2}}{\sqrt{a+b x}} \, dx}{b^2 (b c-a d)^2}\\ &=\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{\left (5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \int \frac{(c+d x)^{3/2}}{\sqrt{a+b x}} \, dx}{6 b^3 (b c-a d)}\\ &=\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{\left (5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \int \frac{\sqrt{c+d x}}{\sqrt{a+b x}} \, dx}{8 b^4}\\ &=\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^5}+\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{\left (5 (b c-a d) \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 b^5}\\ &=\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^5}+\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{\left (5 (b c-a d) \left (b^2 c^2-14 a b c d+21 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^6}\\ &=\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^5}+\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{\left (5 (b c-a d) \left (b^2 c^2-14 a b c d+21 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^6}\\ &=\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 b^5}+\frac{5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac{\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt{a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac{2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac{4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt{a+b x}}+\frac{5 (b c-a d) \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{11/2} \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.807928, size = 218, normalized size = 0.68 \[ \frac{\sqrt{c+d x} \left (\frac{a^2 b^2 \left (113 c^2-574 c d x+63 d^2 x^2\right )+420 a^3 b d (d x-c)+315 a^4 d^2-6 a b^3 x \left (-27 c^2+16 c d x+3 d^2 x^2\right )+b^4 x^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )}{(a+b x)^{3/2}}+\frac{15 \sqrt{b c-a d} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right ) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{d} \sqrt{\frac{b (c+d x)}{b c-a d}}}\right )}{24 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*((315*a^4*d^2 + 420*a^3*b*d*(-c + d*x) - 6*a*b^3*x*(-27*c^2 + 16*c*d*x + 3*d^2*x^2) + b^4*x^2*(
33*c^2 + 26*c*d*x + 8*d^2*x^2) + a^2*b^2*(113*c^2 - 574*c*d*x + 63*d^2*x^2))/(a + b*x)^(3/2) + (15*Sqrt[b*c -
a*d]*(b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[(b*(c
+ d*x))/(b*c - a*d)])))/(24*b^5)

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Maple [B]  time = 0.025, size = 1002, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(d*x+c)^(1/2)*(-16*x^4*b^4*d^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+315*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^3*b^2*d^3-525*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*b^3*c*d^2+225*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^4*c^2*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)
)*x^2*b^5*c^3+36*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^3*a*b^3*d^2-52*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^3*
b^4*c*d+630*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^4*b*d^3-1050*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^3*b^2*c*d^2+450*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*b^3*c^2*d-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*a*b^4*c^3-126*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^2*b^2*d^2+
192*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a*b^3*c*d-66*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*b^4*c^2+315*l
n(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^5*d^3-525*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^4*b*c*d^2+225*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^3*b^2*c^2*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))*a^2*b^3*c^3-840*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^3*b*d^2+1148*(b*d)^(1/2)*((b*x+a)*(d
*x+c))^(1/2)*x*a^2*b^2*c*d-324*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b^3*c^2-630*(b*d)^(1/2)*((b*x+a)*(d*x+c
))^(1/2)*a^4*d^2+840*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*c*d-226*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2
*b^2*c^2)/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(3/2)/b^5

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 10.7029, size = 1770, normalized size = 5.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(15*(a^2*b^3*c^3 - 15*a^3*b^2*c^2*d + 35*a^4*b*c*d^2 - 21*a^5*d^3 + (b^5*c^3 - 15*a*b^4*c^2*d + 35*a^2*
b^3*c*d^2 - 21*a^3*b^2*d^3)*x^2 + 2*(a*b^4*c^3 - 15*a^2*b^3*c^2*d + 35*a^3*b^2*c*d^2 - 21*a^4*b*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*(8*b^5*d^3*x^4 + 113*a^2*b^3*c^2*d - 420*a^3*b^2*c*d^2 + 315*a^4*b*d^3
+ 2*(13*b^5*c*d^2 - 9*a*b^4*d^3)*x^3 + 3*(11*b^5*c^2*d - 32*a*b^4*c*d^2 + 21*a^2*b^3*d^3)*x^2 + 2*(81*a*b^4*c^
2*d - 287*a^2*b^3*c*d^2 + 210*a^3*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^8*d*x^2 + 2*a*b^7*d*x + a^2*b^6*
d), -1/48*(15*(a^2*b^3*c^3 - 15*a^3*b^2*c^2*d + 35*a^4*b*c*d^2 - 21*a^5*d^3 + (b^5*c^3 - 15*a*b^4*c^2*d + 35*a
^2*b^3*c*d^2 - 21*a^3*b^2*d^3)*x^2 + 2*(a*b^4*c^3 - 15*a^2*b^3*c^2*d + 35*a^3*b^2*c*d^2 - 21*a^4*b*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*(8*b^5*d^3*x^4 + 113*a^2*b^3*c^2*d - 420*a^3*b^2*c*d^2 + 315*a^4*b*d^3 + 2*(13*b^5*c*d^
2 - 9*a*b^4*d^3)*x^3 + 3*(11*b^5*c^2*d - 32*a*b^4*c*d^2 + 21*a^2*b^3*d^3)*x^2 + 2*(81*a*b^4*c^2*d - 287*a^2*b^
3*c*d^2 + 210*a^3*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^8*d*x^2 + 2*a*b^7*d*x + a^2*b^6*d)]

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

[Out]

Timed out

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Giac [B]  time = 2.16331, size = 1251, normalized size = 3.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x+c)^(5/2)/(b*x+a)^(5/2),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)*d^2*abs(b)/b^7 + (13*b^21*c*d
^5*abs(b) - 25*a*b^20*d^6*abs(b))/(b^27*d^4)) + 3*(11*b^22*c^2*d^4*abs(b) - 58*a*b^21*c*d^5*abs(b) + 55*a^2*b^
20*d^6*abs(b))/(b^27*d^4)) - 5/16*(sqrt(b*d)*b^3*c^3*abs(b) - 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 35*sqrt(b*d)*a
^2*b*c*d^2*abs(b) - 21*sqrt(b*d)*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a
*b*d))^2)/(b^7*d) + 4/3*(6*sqrt(b*d)*a*b^7*c^5*abs(b) - 37*sqrt(b*d)*a^2*b^6*c^4*d*abs(b) + 88*sqrt(b*d)*a^3*b
^5*c^3*d^2*abs(b) - 102*sqrt(b*d)*a^4*b^4*c^2*d^3*abs(b) + 58*sqrt(b*d)*a^5*b^3*c*d^4*abs(b) - 13*sqrt(b*d)*a^
6*b^2*d^5*abs(b) - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^5*c^4*ab
s(b) + 60*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^4*c^3*d*abs(b) - 1
08*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^3*c^2*d^2*abs(b) + 84*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^2*c*d^3*abs(b) - 24*sqrt(b*d)*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b*d^4*abs(b) + 6*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^3*c^3*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^2*c^2*d*abs(b) + 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^
2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b*c*d^2*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d))^4*a^4*d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^2)^3*b^6)