3.22.17 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=111 \[ -\frac {2 (d+e x)^{3/2} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)}+\frac {2 B \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2}}-\frac {2 B \sqrt {d+e x}}{b^2 \sqrt {a+b x}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {78, 47, 63, 217, 206} \begin {gather*} -\frac {2 (d+e x)^{3/2} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)}-\frac {2 B \sqrt {d+e x}}{b^2 \sqrt {a+b x}}+\frac {2 B \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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

________________________________________________________________________________________

Mathematica [C]  time = 0.17, size = 114, normalized size = 1.03 \begin {gather*} \frac {2 \sqrt {d+e x} \left ((d+e x) (B d-A e)-\frac {B (b d-a e)^2 \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {e (a+b x)}{a e-b d}\right )}{b^2 \sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{3 e (a+b x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*((B*d - A*e)*(d + e*x) - (B*(b*d - a*e)^2*Hypergeometric2F1[-3/2, -3/2, -1/2, (e*(a + b*x))/(
-(b*d) + a*e)])/(b^2*Sqrt[(b*(d + e*x))/(b*d - a*e)])))/(3*e*(b*d - a*e)*(a + b*x)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.22, size = 124, normalized size = 1.12 \begin {gather*} \frac {2 B \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{5/2}}-\frac {2 \sqrt {d+e x} \left (\frac {A b^2 (d+e x)}{a+b x}-\frac {a b B (d+e x)}{a+b x}-3 a B e+3 b B d\right )}{3 b^2 \sqrt {a+b x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 4.34, size = 525, normalized size = 4.73 \begin {gather*} \left [\frac {3 \, {\left (B a^{2} b d - B a^{3} e + {\left (B b^{3} d - B a b^{2} e\right )} x^{2} + 2 \, {\left (B a b^{2} d - B a^{2} b e\right )} x\right )} \sqrt {\frac {e}{b}} \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^{2} e x + b^{2} d + a b e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {e}{b}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (3 \, B a^{2} e - {\left (2 \, B a b + A b^{2}\right )} d - {\left (3 \, B b^{2} d - {\left (4 \, B a b - A b^{2}\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (a^{2} b^{3} d - a^{3} b^{2} e + {\left (b^{5} d - a b^{4} e\right )} x^{2} + 2 \, {\left (a b^{4} d - a^{2} b^{3} e\right )} x\right )}}, -\frac {3 \, {\left (B a^{2} b d - B a^{3} e + {\left (B b^{3} d - B a b^{2} e\right )} x^{2} + 2 \, {\left (B a b^{2} d - B a^{2} b e\right )} x\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {e}{b}}}{2 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )}}\right ) - 2 \, {\left (3 \, B a^{2} e - {\left (2 \, B a b + A b^{2}\right )} d - {\left (3 \, B b^{2} d - {\left (4 \, B a b - A b^{2}\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a^{2} b^{3} d - a^{3} b^{2} e + {\left (b^{5} d - a b^{4} e\right )} x^{2} + 2 \, {\left (a b^{4} d - a^{2} b^{3} e\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.81, size = 522, normalized size = 4.70 \begin {gather*} -\frac {B {\left | b \right |} e^{\frac {1}{2}} \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 )}^{2}\right )}{b^{\frac {7}{2}}} - \frac {4 \, {\left (3 \, B b^{\frac {11}{2}} d^{3} {\left | b \right |} e^{\frac {1}{2}} - 10 \, B a b^{\frac {9}{2}} d^{2} {\left | b \right |} e^{\frac {3}{2}} + A b^{\frac {11}{2}} d^{2} {\left | b \right |} e^{\frac {3}{2}} - 6 \, {\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 )}^{2} B b^{\frac {7}{2}} d^{2} {\left | b \right |} e^{\frac {1}{2}} + 11 \, B a^{2} b^{\frac {7}{2}} d {\left | b \right |} e^{\frac {5}{2}} - 2 \, A a b^{\frac {9}{2}} d {\left | b \right |} e^{\frac {5}{2}} + 12 \, {\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 )}^{2} B a b^{\frac {5}{2}} d {\left | b \right |} e^{\frac {3}{2}} + 3 \, {\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 )}^{4} B b^{\frac {3}{2}} d {\left | b \right |} e^{\frac {1}{2}} - 4 \, B a^{3} b^{\frac {5}{2}} {\left | b \right |} e^{\frac {7}{2}} + A a^{2} b^{\frac {7}{2}} {\left | b \right |} e^{\frac {7}{2}} - 6 \, {\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 )}^{2} B a^{2} b^{\frac {3}{2}} {\left | b \right |} e^{\frac {5}{2}} - 6 \, {\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 )}^{4} B a \sqrt {b} {\left | b \right |} e^{\frac {3}{2}} + 3 \, {\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 )}^{4} A b^{\frac {3}{2}} {\left | b \right |} e^{\frac {3}{2}}\right )}}{3 \, {\left (b^{2} d - a b e - {\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 )}^{2}\right )}^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-B*abs(b)*e^(1/2)*log((sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)/b^(7/2) - 4/3*(
3*B*b^(11/2)*d^3*abs(b)*e^(1/2) - 10*B*a*b^(9/2)*d^2*abs(b)*e^(3/2) + A*b^(11/2)*d^2*abs(b)*e^(3/2) - 6*(sqrt(
b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^(7/2)*d^2*abs(b)*e^(1/2) + 11*B*a^2*b^(7
/2)*d*abs(b)*e^(5/2) - 2*A*a*b^(9/2)*d*abs(b)*e^(5/2) + 12*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x
+ a)*b*e - a*b*e))^2*B*a*b^(5/2)*d*abs(b)*e^(3/2) + 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*
b*e - a*b*e))^4*B*b^(3/2)*d*abs(b)*e^(1/2) - 4*B*a^3*b^(5/2)*abs(b)*e^(7/2) + A*a^2*b^(7/2)*abs(b)*e^(7/2) - 6
*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a^2*b^(3/2)*abs(b)*e^(5/2) - 6*(sqr
t(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*B*a*sqrt(b)*abs(b)*e^(3/2) + 3*(sqrt(b*x +
a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*A*b^(3/2)*abs(b)*e^(3/2))/((b^2*d - a*b*e - (sqrt
(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)^3*b^3)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 503, normalized size = 4.53 \begin {gather*} \frac {\left (3 B a \,b^{2} e^{2} x^{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 )-3 B \,b^{3} d e \,x^{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 )+6 B \,a^{2} b \,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 )-6 B a \,b^{2} d 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 )+3 B \,a^{3} 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 )-3 B \,a^{2} b d 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 )+2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,b^{2} e x -8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b e x +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,b^{2} d -6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,a^{2} e +4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b d \right ) \sqrt {e x +d}}{3 \sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \left (b x +a \right )^{\frac {3}{2}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x)

[Out]

1/3*(3*B*a*b^2*e^2*x^2*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))-3*B*b^3*d*e
*x^2*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))+6*B*a^2*b*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))-6*B*a*b^2*d*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))+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b^2*e*x+3*B*a^3*e^2*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))-3*B*a^2*b*d*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))-8*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*b*e*x+6*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2)*B*b^2*d*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b^2*d-6*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)*B*a^2*e+4*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*b*d)*(e*x+d)^(1/2)/(b*e)^(1/2)/(a*e-b*d)/((b*x+a)
*(e*x+d))^(1/2)/b^2/(b*x+a)^(3/2)

________________________________________________________________________________________

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)*(e*x+d)^(1/2)/(b*x+a)^(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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {d+e\,x}}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(1/2))/(a + b*x)^(5/2),x)

[Out]

int(((A + B*x)*(d + e*x)^(1/2))/(a + b*x)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \sqrt {d + e x}}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(1/2)/(b*x+a)**(5/2),x)

[Out]

Integral((A + B*x)*sqrt(d + e*x)/(a + b*x)**(5/2), x)

________________________________________________________________________________________