3.21.76 \(\int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=202 \[ \frac {3 (b d-a e) (-a B e-4 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 \sqrt {b} e^{7/2}}-\frac {3 \sqrt {a+b x} \sqrt {d+e x} (-a B e-4 A b e+5 b B d)}{4 e^3}+\frac {(a+b x)^{3/2} \sqrt {d+e x} (-a B e-4 A b e+5 b B d)}{2 e^2 (b d-a e)}-\frac {2 (a+b x)^{5/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)} \]

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Rubi [A]  time = 0.17, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {78, 50, 63, 217, 206} \begin {gather*} \frac {(a+b x)^{3/2} \sqrt {d+e x} (-a B e-4 A b e+5 b B d)}{2 e^2 (b d-a e)}-\frac {3 \sqrt {a+b x} \sqrt {d+e x} (-a B e-4 A b e+5 b B d)}{4 e^3}+\frac {3 (b d-a e) (-a B e-4 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 \sqrt {b} e^{7/2}}-\frac {2 (a+b x)^{5/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 {(a+b x)^{3/2} (A+B x)}{(d+e x)^{3/2}} \, dx &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{e (b d-a e) \sqrt {d+e x}}+\frac {(5 b B d-4 A b e-a B e) \int \frac {(a+b x)^{3/2}}{\sqrt {d+e x}} \, dx}{e (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{e (b d-a e) \sqrt {d+e x}}+\frac {(5 b B d-4 A b e-a B e) (a+b x)^{3/2} \sqrt {d+e x}}{2 e^2 (b d-a e)}-\frac {(3 (5 b B d-4 A b e-a B e)) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}} \, dx}{4 e^2}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{e (b d-a e) \sqrt {d+e x}}-\frac {3 (5 b B d-4 A b e-a B e) \sqrt {a+b x} \sqrt {d+e x}}{4 e^3}+\frac {(5 b B d-4 A b e-a B e) (a+b x)^{3/2} \sqrt {d+e x}}{2 e^2 (b d-a e)}+\frac {(3 (b d-a e) (5 b B d-4 A b e-a B e)) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{8 e^3}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{e (b d-a e) \sqrt {d+e x}}-\frac {3 (5 b B d-4 A b e-a B e) \sqrt {a+b x} \sqrt {d+e x}}{4 e^3}+\frac {(5 b B d-4 A b e-a B e) (a+b x)^{3/2} \sqrt {d+e x}}{2 e^2 (b d-a e)}+\frac {(3 (b d-a e) (5 b B d-4 A b e-a B e)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d-\frac {a e}{b}+\frac {e x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{4 b e^3}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{e (b d-a e) \sqrt {d+e x}}-\frac {3 (5 b B d-4 A b e-a B e) \sqrt {a+b x} \sqrt {d+e x}}{4 e^3}+\frac {(5 b B d-4 A b e-a B e) (a+b x)^{3/2} \sqrt {d+e x}}{2 e^2 (b d-a e)}+\frac {(3 (b d-a e) (5 b B d-4 A b e-a B e)) \operatorname {Subst}\left (\int \frac {1}{1-\frac {e x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {d+e x}}\right )}{4 b e^3}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{e (b d-a e) \sqrt {d+e x}}-\frac {3 (5 b B d-4 A b e-a B e) \sqrt {a+b x} \sqrt {d+e x}}{4 e^3}+\frac {(5 b B d-4 A b e-a B e) (a+b x)^{3/2} \sqrt {d+e x}}{2 e^2 (b d-a e)}+\frac {3 (b d-a e) (5 b B d-4 A b e-a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 \sqrt {b} e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 168, normalized size = 0.83 \begin {gather*} \frac {\sqrt {e} \sqrt {a+b x} \left (a e (-8 A e+13 B d+5 B e x)+4 A b e (3 d+e x)+b B \left (-15 d^2-5 d e x+2 e^2 x^2\right )\right )+\frac {3 (b d-a e)^{3/2} \sqrt {\frac {b (d+e x)}{b d-a e}} (-a B e-4 A b e+5 b B d) \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{b}}{4 e^{7/2} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*Sqrt[a + b*x]*(4*A*b*e*(3*d + e*x) + a*e*(13*B*d - 8*A*e + 5*B*e*x) + b*B*(-15*d^2 - 5*d*e*x + 2*e^2*
x^2)) + (3*(b*d - a*e)^(3/2)*(5*b*B*d - 4*A*b*e - a*B*e)*Sqrt[(b*(d + e*x))/(b*d - a*e)]*ArcSinh[(Sqrt[e]*Sqrt
[a + b*x])/Sqrt[b*d - a*e]])/b)/(4*e^(7/2)*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 1.57, size = 214, normalized size = 1.06 \begin {gather*} \frac {3 \sqrt {\frac {b}{e}} \left (-a^2 B e^2-4 a A b e^2+6 a b B d e+4 A b^2 d e-5 b^2 B d^2\right ) \log \left (\sqrt {a+\frac {b (d+e x)}{e}-\frac {b d}{e}}-\sqrt {\frac {b}{e}} \sqrt {d+e x}\right )}{4 b e^3}+\frac {\sqrt {a+\frac {b (d+e x)}{e}-\frac {b d}{e}} \left (-8 a A e^2+5 a B e (d+e x)+8 a B d e+4 A b e (d+e x)+8 A b d e-8 b B d^2-9 b B d (d+e x)+2 b B (d+e x)^2\right )}{4 e^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - (b*d)/e + (b*(d + e*x))/e]*(-8*b*B*d^2 + 8*A*b*d*e + 8*a*B*d*e - 8*a*A*e^2 - 9*b*B*d*(d + e*x) + 4*A
*b*e*(d + e*x) + 5*a*B*e*(d + e*x) + 2*b*B*(d + e*x)^2))/(4*e^3*Sqrt[d + e*x]) + (3*Sqrt[b/e]*(-5*b^2*B*d^2 +
4*A*b^2*d*e + 6*a*b*B*d*e - 4*a*A*b*e^2 - a^2*B*e^2)*Log[-(Sqrt[b/e]*Sqrt[d + e*x]) + Sqrt[a - (b*d)/e + (b*(d
+ e*x))/e]])/(4*b*e^3)

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fricas [A]  time = 3.14, size = 572, normalized size = 2.83 \begin {gather*} \left [\frac {3 \, {\left (5 \, B b^{2} d^{3} - 2 \, {\left (3 \, B a b + 2 \, A b^{2}\right )} d^{2} e + {\left (B a^{2} + 4 \, A a b\right )} d e^{2} + {\left (5 \, B b^{2} d^{2} e - 2 \, {\left (3 \, B a b + 2 \, A b^{2}\right )} d e^{2} + {\left (B a^{2} + 4 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (2 \, B b^{2} e^{3} x^{2} - 15 \, B b^{2} d^{2} e - 8 \, A a b e^{3} + {\left (13 \, B a b + 12 \, A b^{2}\right )} d e^{2} - {\left (5 \, B b^{2} d e^{2} - {\left (5 \, B a b + 4 \, A b^{2}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{16 \, {\left (b e^{5} x + b d e^{4}\right )}}, -\frac {3 \, {\left (5 \, B b^{2} d^{3} - 2 \, {\left (3 \, B a b + 2 \, A b^{2}\right )} d^{2} e + {\left (B a^{2} + 4 \, A a b\right )} d e^{2} + {\left (5 \, B b^{2} d^{2} e - 2 \, {\left (3 \, B a b + 2 \, A b^{2}\right )} d e^{2} + {\left (B a^{2} + 4 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, B b^{2} e^{3} x^{2} - 15 \, B b^{2} d^{2} e - 8 \, A a b e^{3} + {\left (13 \, B a b + 12 \, A b^{2}\right )} d e^{2} - {\left (5 \, B b^{2} d e^{2} - {\left (5 \, B a b + 4 \, A b^{2}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{8 \, {\left (b e^{5} x + b d e^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*(5*B*b^2*d^3 - 2*(3*B*a*b + 2*A*b^2)*d^2*e + (B*a^2 + 4*A*a*b)*d*e^2 + (5*B*b^2*d^2*e - 2*(3*B*a*b +
2*A*b^2)*d*e^2 + (B*a^2 + 4*A*a*b)*e^3)*x)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*
b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) + 4*(2*B*b^2*e^3*x^2 - 15*
B*b^2*d^2*e - 8*A*a*b*e^3 + (13*B*a*b + 12*A*b^2)*d*e^2 - (5*B*b^2*d*e^2 - (5*B*a*b + 4*A*b^2)*e^3)*x)*sqrt(b*
x + a)*sqrt(e*x + d))/(b*e^5*x + b*d*e^4), -1/8*(3*(5*B*b^2*d^3 - 2*(3*B*a*b + 2*A*b^2)*d^2*e + (B*a^2 + 4*A*a
*b)*d*e^2 + (5*B*b^2*d^2*e - 2*(3*B*a*b + 2*A*b^2)*d*e^2 + (B*a^2 + 4*A*a*b)*e^3)*x)*sqrt(-b*e)*arctan(1/2*(2*
b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) - 2
*(2*B*b^2*e^3*x^2 - 15*B*b^2*d^2*e - 8*A*a*b*e^3 + (13*B*a*b + 12*A*b^2)*d*e^2 - (5*B*b^2*d*e^2 - (5*B*a*b + 4
*A*b^2)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*e^5*x + b*d*e^4)]

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giac [A]  time = 1.64, size = 257, normalized size = 1.27 \begin {gather*} -\frac {3 \, {\left (5 \, B b^{2} d^{2} {\left | b \right |} - 6 \, B a b d {\left | b \right |} e - 4 \, A b^{2} d {\left | b \right |} e + B a^{2} {\left | b \right |} e^{2} + 4 \, A a b {\left | b \right |} e^{2}\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 )}{4 \, b^{\frac {3}{2}}} + \frac {{\left ({\left (\frac {2 \, {\left (b x + a\right )} B {\left | b \right |} e^{\left (-1\right )}}{b} - \frac {{\left (5 \, B b^{2} d {\left | b \right |} e^{3} - B a b {\left | b \right |} e^{4} - 4 \, A b^{2} {\left | b \right |} e^{4}\right )} e^{\left (-5\right )}}{b^{2}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (5 \, B b^{3} d^{2} {\left | b \right |} e^{2} - 6 \, B a b^{2} d {\left | b \right |} e^{3} - 4 \, A b^{3} d {\left | b \right |} e^{3} + B a^{2} b {\left | b \right |} e^{4} + 4 \, A a b^{2} {\left | b \right |} e^{4}\right )} e^{\left (-5\right )}}{b^{2}}\right )} \sqrt {b x + a}}{4 \, \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} \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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 740, normalized size = 3.66 \begin {gather*} \frac {\sqrt {b x +a}\, \left (12 A a b \,e^{3} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-12 A \,b^{2} d \,e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+3 B \,a^{2} e^{3} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-18 B a b d \,e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+15 B \,b^{2} d^{2} e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+12 A a b d \,e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-12 A \,b^{2} d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+3 B \,a^{2} d \,e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-18 B a b \,d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+15 B \,b^{2} d^{3} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b \,e^{2} x^{2}+8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b \,e^{2} x +10 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a \,e^{2} x -10 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b d e x -16 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A a \,e^{2}+24 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b d e +26 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a d e -30 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b \,d^{2}\right )}{8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, \sqrt {e x +d}\, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(12*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*a*b*e^3-
12*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*b^2*d*e^2+3*B*ln(1/2*(2*b*e
*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x*a^2*e^3-18*B*a*b*d*e^2*x*ln(1/2*(2*b*e*x+a*e+
b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+15*B*b^2*d^2*e*x*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e
*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+4*B*x^2*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+12*A*ln(1/2*(2*b*e*x+
a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a*b*d*e^2-12*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*b^2*d^2*e+8*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b*e^2*x+3*B*ln(1/2*
(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a^2*d*e^2-18*B*a*b*d^2*e*ln(1/2*(2*b*e*x+
a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+15*B*b^2*d^3*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e
*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+10*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*e^2*x-10*((b*x+a)*(e*x+d))^(
1/2)*(b*e)^(1/2)*B*b*d*e*x-16*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*a*e^2+24*A*b*d*e*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)+26*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*d*e-30*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b*d^2)/((b*
x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)/(e*x+d)^(1/2)/e^3

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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)^(3/2),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*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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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 )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (a + b x\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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