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

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

[Out]

1/64*(-a*d+b*c)*(5*a^3*d^3+9*a^2*b*c*d^2+15*a*b^2*c^2*d+35*b^3*c^3)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x
+c)^(1/2))/b^(7/2)/d^(9/2)+1/4*x^2*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b/d+1/96*(b*x+a)^(3/2)*(35*b^2*c^2+22*a*b*c*d+1
5*a^2*d^2-4*b*d*(5*a*d+7*b*c)*x)*(d*x+c)^(1/2)/b^3/d^3-1/64*(5*a^3*d^3+9*a^2*b*c*d^2+15*a*b^2*c^2*d+35*b^3*c^3
)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^4

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Rubi [A]  time = 0.17, antiderivative size = 251, normalized size of antiderivative = 1.00, 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 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 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 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^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*}

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Mathematica [A]  time = 0.90, size = 227, normalized size = 0.90 \[ \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) \left (15 a^3 d^3+a^2 b d^2 (17 c-10 d x)+a b^2 d \left (25 c^2-12 c d x+8 d^2 x^2\right )+b^3 \left (-105 c^3+70 c^2 d x-56 c d^2 x^2+48 d^3 x^3\right )\right )+3 \left (5 a^3 d^3+9 a^2 b c d^2+15 a b^2 c^2 d+35 b^3 c^3\right ) (b c-a d)^{3/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{192 b^4 d^{9/2} \sqrt {c+d x}} \]

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])

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fricas [A]  time = 1.07, size = 546, normalized size = 2.18 \[ \left [-\frac {3 \, {\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 25 \, a b^{3} c^{2} d^{2} + 17 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 6 \, a b^{3} c d^{3} - 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{4} d^{5}}, -\frac {3 \, {\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 25 \, a b^{3} c^{2} d^{2} + 17 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 6 \, a b^{3} c d^{3} - 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{4} d^{5}}\right ] \]

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)]

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giac [A]  time = 2.02, size = 291, normalized size = 1.16 \[ \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{4} d} - \frac {7 \, b^{13} c d^{5} + 17 \, a b^{12} d^{6}}{b^{16} d^{7}}\right )} + \frac {35 \, b^{14} c^{2} d^{4} + 50 \, a b^{13} c d^{5} + 59 \, a^{2} b^{12} d^{6}}{b^{16} d^{7}}\right )} - \frac {3 \, {\left (35 \, b^{15} c^{3} d^{3} + 15 \, a b^{14} c^{2} d^{4} + 9 \, a^{2} b^{13} c d^{5} + 5 \, a^{3} b^{12} d^{6}\right )}}{b^{16} d^{7}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{3} d^{4}}\right )} b}{192 \, {\left | b \right |}} \]

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)

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maple [B]  time = 0.03, size = 574, normalized size = 2.29 \[ -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (15 a^{4} d^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+12 a^{3} b c \,d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+18 a^{2} b^{2} c^{2} d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+60 a \,b^{3} c^{3} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-105 b^{4} c^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-96 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} d^{3} x^{3}-16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} d^{3} x^{2}+112 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c \,d^{2} x^{2}+20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b \,d^{3} x +24 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c \,d^{2} x -140 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c^{2} d x -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} d^{3}-34 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b c \,d^{2}-50 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c^{2} d +210 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{3} c^{3}\right )}{384 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} d^{4}} \]

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*x+a)*(
d*x+c))^(1/2)*(b*d)^(1/2)*x*a^2*b*d^3+24*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*a*b^2*c*d^2-140*((b*x+a)*(d*x+c
))^(1/2)*(b*d)^(1/2)*x*b^3*c^2*d-30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^3*d^3-34*((b*x+a)*(d*x+c))^(1/2)*(b*
d)^(1/2)*a^2*b*c*d^2-50*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(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)

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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^3*(b*x+a)^(1/2)/(d*x+c)^(1/2),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(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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

[Out]

Timed out

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