3.23.16 \(\int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{5/2}} \, dx\) [2216]

Optimal. Leaf size=202 \[ -\frac {2 (B d-A e) (a+b x)^{5/2}}{3 e (b d-a e) (d+e x)^{3/2}}-\frac {2 (5 b B d-2 A b e-3 a B e) (a+b x)^{3/2}}{3 e^2 (b d-a e) \sqrt {d+e x}}+\frac {b (5 b B d-2 A b e-3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^3 (b d-a e)}-\frac {\sqrt {b} (5 b B d-2 A b e-3 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{7/2}} \]

[Out]

-2/3*(-A*e+B*d)*(b*x+a)^(5/2)/e/(-a*e+b*d)/(e*x+d)^(3/2)-(-2*A*b*e-3*B*a*e+5*B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1
/2)/b^(1/2)/(e*x+d)^(1/2))*b^(1/2)/e^(7/2)-2/3*(-2*A*b*e-3*B*a*e+5*B*b*d)*(b*x+a)^(3/2)/e^2/(-a*e+b*d)/(e*x+d)
^(1/2)+b*(-2*A*b*e-3*B*a*e+5*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/e^3/(-a*e+b*d)

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Rubi [A]
time = 0.10, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 49, 52, 65, 223, 212} \begin {gather*} -\frac {\sqrt {b} (-3 a B e-2 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{7/2}}+\frac {b \sqrt {a+b x} \sqrt {d+e x} (-3 a B e-2 A b e+5 b B d)}{e^3 (b d-a e)}-\frac {2 (a+b x)^{3/2} (-3 a B e-2 A b e+5 b B d)}{3 e^2 \sqrt {d+e x} (b d-a e)}-\frac {2 (a+b x)^{5/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(3*e*(b*d - a*e)*(d + e*x)^(3/2)) - (2*(5*b*B*d - 2*A*b*e - 3*a*B*e)*(a + b*x
)^(3/2))/(3*e^2*(b*d - a*e)*Sqrt[d + e*x]) + (b*(5*b*B*d - 2*A*b*e - 3*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(e^
3*(b*d - a*e)) - (Sqrt[b]*(5*b*B*d - 2*A*b*e - 3*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x]
)])/e^(7/2)

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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

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Mathematica [A]
time = 0.33, size = 136, normalized size = 0.67 \begin {gather*} \frac {\sqrt {a+b x} \left (-2 A b e (3 d+4 e x)-2 a e (2 B d+A e+3 B e x)+b B \left (15 d^2+20 d e x+3 e^2 x^2\right )\right )}{3 e^3 (d+e x)^{3/2}}+\frac {\sqrt {b} (-5 b B d+2 A b e+3 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-2*A*b*e*(3*d + 4*e*x) - 2*a*e*(2*B*d + A*e + 3*B*e*x) + b*B*(15*d^2 + 20*d*e*x + 3*e^2*x^2)))
/(3*e^3*(d + e*x)^(3/2)) + (Sqrt[b]*(-5*b*B*d + 2*A*b*e + 3*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sq
rt[a + b*x])])/e^(7/2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(697\) vs. \(2(174)=348\).
time = 0.11, size = 698, normalized size = 3.46

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (6 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} e^{3} x^{2}+9 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b \,e^{3} x^{2}-15 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d \,e^{2} x^{2}+12 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d \,e^{2} x +18 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b d \,e^{2} x -30 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d^{2} e x +6 B b \,e^{2} x^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+6 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d^{2} e -16 A b \,e^{2} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+9 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b \,d^{2} e -15 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d^{3}-12 B a \,e^{2} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+40 B b d e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-4 A a \,e^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-12 A b d e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-8 B a d e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+30 B b \,d^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right )}{6 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, e^{3} \left (e x +d \right )^{\frac {3}{2}}}\)	\(698\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(b*x+a)^(1/2)*(6*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*e^3*x^2
+9*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*e^3*x^2-15*B*ln(1/2*(2*b*
e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d*e^2*x^2+12*A*ln(1/2*(2*b*e*x+2*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d*e^2*x+18*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b
*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d*e^2*x-30*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d
)/(b*e)^(1/2))*b^2*d^2*e*x+6*B*b*e^2*x^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e
*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^2*e-16*A*b*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+9*B*
ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d^2*e-15*B*ln(1/2*(2*b*e*x+2*(
(b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^3-12*B*a*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1
/2)+40*B*b*d*e*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-4*A*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-12*A*b*d*e*
((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-8*B*a*d*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+30*B*b*d^2*((b*x+a)*(e*x+d))
^(1/2)*(b*e)^(1/2))/(b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/e^3/(e*x+d)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.55, size = 514, normalized size = 2.54 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b d^{3} - {\left (3 \, B a + 2 \, A b\right )} x^{2} e^{3} + {\left (5 \, B b d x^{2} - 2 \, {\left (3 \, B a + 2 \, A b\right )} d x\right )} e^{2} + {\left (10 \, B b d^{2} x - {\left (3 \, B a + 2 \, A b\right )} d^{2}\right )} e\right )} \sqrt {b} e^{\left (-\frac {1}{2}\right )} \log \left (b^{2} d^{2} + 4 \, {\left (b d e + {\left (2 \, b x + a\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\left (-\frac {1}{2}\right )} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) - 4 \, {\left (15 \, B b d^{2} + {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} e^{2} + 2 \, {\left (10 \, B b d x - {\left (2 \, B a + 3 \, A b\right )} d\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{12 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}}, \frac {3 \, {\left (5 \, B b d^{3} - {\left (3 \, B a + 2 \, A b\right )} x^{2} e^{3} + {\left (5 \, B b d x^{2} - 2 \, {\left (3 \, B a + 2 \, A b\right )} d x\right )} e^{2} + {\left (10 \, B b d^{2} x - {\left (3 \, B a + 2 \, A b\right )} d^{2}\right )} e\right )} \sqrt {-b e^{\left (-1\right )}} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {-b e^{\left (-1\right )}}}{2 \, {\left (b^{2} d x + a b d + {\left (b^{2} x^{2} + a b x\right )} e\right )}}\right ) + 2 \, {\left (15 \, B b d^{2} + {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} e^{2} + 2 \, {\left (10 \, B b d x - {\left (2 \, B a + 3 \, A b\right )} d\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{6 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*x^2*e^3 + (5*B*b*d*x^2 - 2*(3*B*a + 2*A*b)*d*x)*e^2 + (10*B*b*d^2*x - (
3*B*a + 2*A*b)*d^2)*e)*sqrt(b)*e^(-1/2)*log(b^2*d^2 + 4*(b*d*e + (2*b*x + a)*e^2)*sqrt(b*x + a)*sqrt(x*e + d)*
sqrt(b)*e^(-1/2) + (8*b^2*x^2 + 8*a*b*x + a^2)*e^2 + 2*(4*b^2*d*x + 3*a*b*d)*e) - 4*(15*B*b*d^2 + (3*B*b*x^2 -
2*A*a - 2*(3*B*a + 4*A*b)*x)*e^2 + 2*(10*B*b*d*x - (2*B*a + 3*A*b)*d)*e)*sqrt(b*x + a)*sqrt(x*e + d))/(x^2*e^
5 + 2*d*x*e^4 + d^2*e^3), 1/6*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*x^2*e^3 + (5*B*b*d*x^2 - 2*(3*B*a + 2*A*b)*d*x)*
e^2 + (10*B*b*d^2*x - (3*B*a + 2*A*b)*d^2)*e)*sqrt(-b*e^(-1))*arctan(1/2*(b*d + (2*b*x + a)*e)*sqrt(b*x + a)*s
qrt(x*e + d)*sqrt(-b*e^(-1))/(b^2*d*x + a*b*d + (b^2*x^2 + a*b*x)*e)) + 2*(15*B*b*d^2 + (3*B*b*x^2 - 2*A*a - 2
*(3*B*a + 4*A*b)*x)*e^2 + 2*(10*B*b*d*x - (2*B*a + 3*A*b)*d)*e)*sqrt(b*x + a)*sqrt(x*e + d))/(x^2*e^5 + 2*d*x*
e^4 + d^2*e^3)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 1.66, size = 352, normalized size = 1.74 \begin {gather*} \frac {{\left (5 \, B b d {\left | b \right |} - 3 \, B a {\left | b \right |} e - 2 \, A b {\left | b \right |} e\right )} e^{\left (-\frac {7}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} + \frac {{\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (B b^{5} d {\left | b \right |} e^{4} - B a b^{4} {\left | b \right |} e^{5}\right )} {\left (b x + a\right )}}{b^{4} d e^{5} - a b^{3} e^{6}} + \frac {4 \, {\left (5 \, B b^{6} d^{2} {\left | b \right |} e^{3} - 8 \, B a b^{5} d {\left | b \right |} e^{4} - 2 \, A b^{6} d {\left | b \right |} e^{4} + 3 \, B a^{2} b^{4} {\left | b \right |} e^{5} + 2 \, A a b^{5} {\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} + \frac {3 \, {\left (5 \, B b^{7} d^{3} {\left | b \right |} e^{2} - 13 \, B a b^{6} d^{2} {\left | b \right |} e^{3} - 2 \, A b^{7} d^{2} {\left | b \right |} e^{3} + 11 \, B a^{2} b^{5} d {\left | b \right |} e^{4} + 4 \, A a b^{6} d {\left | b \right |} e^{4} - 3 \, B a^{3} b^{4} {\left | b \right |} e^{5} - 2 \, A a^{2} b^{5} {\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*B*b*d*abs(b) - 3*B*a*abs(b)*e - 2*A*b*abs(b)*e)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*
d + (b*x + a)*b*e - a*b*e)))/sqrt(b) + 1/3*((b*x + a)*(3*(B*b^5*d*abs(b)*e^4 - B*a*b^4*abs(b)*e^5)*(b*x + a)/(
b^4*d*e^5 - a*b^3*e^6) + 4*(5*B*b^6*d^2*abs(b)*e^3 - 8*B*a*b^5*d*abs(b)*e^4 - 2*A*b^6*d*abs(b)*e^4 + 3*B*a^2*b
^4*abs(b)*e^5 + 2*A*a*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6)) + 3*(5*B*b^7*d^3*abs(b)*e^2 - 13*B*a*b^6*d^2*ab
s(b)*e^3 - 2*A*b^7*d^2*abs(b)*e^3 + 11*B*a^2*b^5*d*abs(b)*e^4 + 4*A*a*b^6*d*abs(b)*e^4 - 3*B*a^3*b^4*abs(b)*e^
5 - 2*A*a^2*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e - a*b*e)^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(3/2))/(d + e*x)^(5/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(3/2))/(d + e*x)^(5/2), x)

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