3.576 \(\int \frac{x^3 \sqrt{a+b x}}{\sqrt{c+d x}} \, dx\)

Optimal. Leaf size=251 \[ \frac{(a+b x)^{3/2} \sqrt{c+d x} \left (15 a^2 d^2-4 b d x (5 a d+7 b c)+22 a b c d+35 b^2 c^2\right )}{96 b^3 d^3}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (9 a^2 b c d^2+5 a^3 d^3+15 a b^2 c^2 d+35 b^3 c^3\right )}{64 b^3 d^4}+\frac{(b c-a d) \left (9 a^2 b c d^2+5 a^3 d^3+15 a b^2 c^2 d+35 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{9/2}}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d} \]

[Out]

-((35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3*d^4) + (x^2*(
a + b*x)^(3/2)*Sqrt[c + d*x])/(4*b*d) + ((a + b*x)^(3/2)*Sqrt[c + d*x]*(35*b^2*c^2 + 22*a*b*c*d + 15*a^2*d^2 -
4*b*d*(7*b*c + 5*a*d)*x))/(96*b^3*d^3) + ((b*c - a*d)*(35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^
3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.16993, antiderivative size = 251, 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 = {100, 147, 50, 63, 217, 206} \[ \frac{(a+b x)^{3/2} \sqrt{c+d x} \left (15 a^2 d^2-4 b d x (5 a d+7 b c)+22 a b c d+35 b^2 c^2\right )}{96 b^3 d^3}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (9 a^2 b c d^2+5 a^3 d^3+15 a b^2 c^2 d+35 b^3 c^3\right )}{64 b^3 d^4}+\frac{(b c-a d) \left (9 a^2 b c d^2+5 a^3 d^3+15 a b^2 c^2 d+35 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{9/2}}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3*d^4) + (x^2*(
a + b*x)^(3/2)*Sqrt[c + d*x])/(4*b*d) + ((a + b*x)^(3/2)*Sqrt[c + d*x]*(35*b^2*c^2 + 22*a*b*c*d + 15*a^2*d^2 -
4*b*d*(7*b*c + 5*a*d)*x))/(96*b^3*d^3) + ((b*c - a*d)*(35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^
3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(9/2))

Rule 100

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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^3 \sqrt{a+b x}}{\sqrt{c+d x}} \, dx &=\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d}+\frac{\int \frac{x \sqrt{a+b x} \left (-2 a c+\frac{1}{2} (-7 b c-5 a d) x\right )}{\sqrt{c+d x}} \, dx}{4 b d}\\ &=\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (35 b^2 c^2+22 a b c d+15 a^2 d^2-4 b d (7 b c+5 a d) x\right )}{96 b^3 d^3}-\frac{\left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{64 b^3 d^3}\\ &=-\frac{\left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^4}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (35 b^2 c^2+22 a b c d+15 a^2 d^2-4 b d (7 b c+5 a d) x\right )}{96 b^3 d^3}+\frac{\left ((b c-a d) \left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{128 b^3 d^4}\\ &=-\frac{\left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^4}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (35 b^2 c^2+22 a b c d+15 a^2 d^2-4 b d (7 b c+5 a d) x\right )}{96 b^3 d^3}+\frac{\left ((b c-a d) \left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\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 )}{64 b^4 d^4}\\ &=-\frac{\left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^4}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (35 b^2 c^2+22 a b c d+15 a^2 d^2-4 b d (7 b c+5 a d) x\right )}{96 b^3 d^3}+\frac{\left ((b c-a d) \left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\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 )}{64 b^4 d^4}\\ &=-\frac{\left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^4}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (35 b^2 c^2+22 a b c d+15 a^2 d^2-4 b d (7 b c+5 a d) x\right )}{96 b^3 d^3}+\frac{(b c-a d) \left (35 b^3 c^3+15 a b^2 c^2 d+9 a^2 b c d^2+5 a^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.861352, size = 227, normalized size = 0.9 \[ \frac{b \sqrt{d} \sqrt{a+b x} (c+d x) \left (a^2 b d^2 (17 c-10 d x)+15 a^3 d^3+a b^2 d \left (25 c^2-12 c d x+8 d^2 x^2\right )+b^3 \left (70 c^2 d x-105 c^3-56 c d^2 x^2+48 d^3 x^3\right )\right )+3 \left (9 a^2 b c d^2+5 a^3 d^3+15 a b^2 c^2 d+35 b^3 c^3\right ) (b c-a d)^{3/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{192 b^4 d^{9/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(15*a^3*d^3 + a^2*b*d^2*(17*c - 10*d*x) + a*b^2*d*(25*c^2 - 12*c*d*x + 8*d^
2*x^2) + b^3*(-105*c^3 + 70*c^2*d*x - 56*c*d^2*x^2 + 48*d^3*x^3)) + 3*(b*c - a*d)^(3/2)*(35*b^3*c^3 + 15*a*b^2
*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c -
a*d]])/(192*b^4*d^(9/2)*Sqrt[c + d*x])

________________________________________________________________________________________

Maple [B]  time = 0.021, size = 574, normalized size = 2.3 \begin{align*} -{\frac{1}{384\,{d}^{4}{b}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( -96\,{x}^{3}{b}^{3}{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-16\,{x}^{2}a{b}^{2}{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+112\,{x}^{2}{b}^{3}c{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+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 ){a}^{4}{d}^{4}+12\,\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}^{3}bc{d}^{3}+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 ){a}^{2}{b}^{2}{c}^{2}{d}^{2}+60\,\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{b}^{3}{c}^{3}d-105\,\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}^{4}{c}^{4}+20\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }x{a}^{2}b{d}^{3}+24\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }xa{b}^{2}c{d}^{2}-140\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }x{b}^{3}{c}^{2}d-30\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{a}^{3}{d}^{3}-34\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{a}^{2}bc{d}^{2}-50\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }a{b}^{2}{c}^{2}d+210\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{b}^{3}{c}^{3} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x+a)^(1/2)/(d*x+c)^(1/2),x)

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*x^3*b^3*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-16*x^2*a*b^2*d^3*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+112*x^2*b^3*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+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^4*d^4+12*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*c*d^3+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^2*b^2*c^2*d^2+60*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
^3*c^3*d-105*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^4*c^4+20*(b*d)^(1/2
)*((b*x+a)*(d*x+c))^(1/2)*x*a^2*b*d^3+24*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b^2*c*d^2-140*(b*d)^(1/2)*((b
*x+a)*(d*x+c))^(1/2)*x*b^3*c^2*d-30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*d^3-34*(b*d)^(1/2)*((b*x+a)*(d*x+c
))^(1/2)*a^2*b*c*d^2-50*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^2*d+210*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2
)*b^3*c^3)/((b*x+a)*(d*x+c))^(1/2)/d^4/b^3/(b*d)^(1/2)

________________________________________________________________________________________

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^3*(b*x+a)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.28905, size = 1216, normalized size = 4.84 \begin{align*} \left [-\frac{3 \,{\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 25 \, a b^{3} c^{2} d^{2} + 17 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \,{\left (7 \, b^{4} c d^{3} - a b^{3} d^{4}\right )} x^{2} + 2 \,{\left (35 \, b^{4} c^{2} d^{2} - 6 \, a b^{3} c d^{3} - 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{768 \, b^{4} d^{5}}, -\frac{3 \,{\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 25 \, a b^{3} c^{2} d^{2} + 17 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \,{\left (7 \, b^{4} c d^{3} - a b^{3} d^{4}\right )} x^{2} + 2 \,{\left (35 \, b^{4} c^{2} d^{2} - 6 \, a b^{3} c d^{3} - 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{384 \, b^{4} d^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/768*(3*(35*b^4*c^4 - 20*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - 5*a^4*d^4)*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*(48*b^4*d^4*x^3 - 105*b^4*c^3*d + 25*a*b^3*c^2*d^2 + 17*a^2*b^2*c*d^3 + 15*a^3*b*d^4 -
8*(7*b^4*c*d^3 - a*b^3*d^4)*x^2 + 2*(35*b^4*c^2*d^2 - 6*a*b^3*c*d^3 - 5*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*
x + c))/(b^4*d^5), -1/384*(3*(35*b^4*c^4 - 20*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - 5*a^4*d^4)*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*(48*b^4*d^4*x^3 - 105*b^4*c^3*d + 25*a*b^3*c^2*d^2 + 17*a^2*b^2*c*d^3 + 15*a^3*b*d^4 -
8*(7*b^4*c*d^3 - a*b^3*d^4)*x^2 + 2*(35*b^4*c^2*d^2 - 6*a*b^3*c*d^3 - 5*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x
+ c))/(b^4*d^5)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.3241, size = 393, normalized size = 1.57 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{6 \,{\left (b x + a\right )}}{b^{4} d} - \frac{7 \, b^{13} c d^{5} + 17 \, a b^{12} d^{6}}{b^{16} d^{7}}\right )} + \frac{35 \, b^{14} c^{2} d^{4} + 50 \, a b^{13} c d^{5} + 59 \, a^{2} b^{12} d^{6}}{b^{16} d^{7}}\right )} - \frac{3 \,{\left (35 \, b^{15} c^{3} d^{3} + 15 \, a b^{14} c^{2} d^{4} + 9 \, a^{2} b^{13} c d^{5} + 5 \, a^{3} b^{12} d^{6}\right )}}{b^{16} d^{7}}\right )} \sqrt{b x + a} - \frac{3 \,{\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\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^{3} d^{4}}\right )} b}{192 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/192*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/(b^4*d) - (7*b^13*c*d^5 + 17
*a*b^12*d^6)/(b^16*d^7)) + (35*b^14*c^2*d^4 + 50*a*b^13*c*d^5 + 59*a^2*b^12*d^6)/(b^16*d^7)) - 3*(35*b^15*c^3*
d^3 + 15*a*b^14*c^2*d^4 + 9*a^2*b^13*c*d^5 + 5*a^3*b^12*d^6)/(b^16*d^7))*sqrt(b*x + a) - 3*(35*b^4*c^4 - 20*a*
b^3*c^3*d - 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - 5*a^4*d^4)*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^3*d^4))*b/abs(b)