3.23.56 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx\) [2256]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {79, 49, 65, 223, 212} \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 {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}} \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 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 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) \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) \text {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) \text {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*}

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Mathematica [A]
time = 0.19, size = 116, normalized size = 1.05 \begin {gather*} -\frac {2 \sqrt {d+e x} \left (A b^2 (d+e x)+B \left (-3 a^2 e+3 b^2 d x+2 a b (d-2 e x)\right )\right )}{3 b^2 (b d-a e) (a+b x)^{3/2}}+\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}} \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]*(A*b^2*(d + e*x) + B*(-3*a^2*e + 3*b^2*d*x + 2*a*b*(d - 2*e*x))))/(3*b^2*(b*d - a*e)*(a + b*
x)^(3/2)) + (2*B*Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/b^(5/2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(502\) vs. \(2(89)=178\).
time = 0.23, size = 503, normalized size = 4.53

method	result	size
default	\(\frac {\left (3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} e^{2} x^{2}-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d e \,x^{2}+6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} b \,e^{2} x -6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d e x +2 A \,b^{2} e x \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}+3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{3} e^{2}-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} b d e -8 B a b e x \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}+6 B \,b^{2} d x \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}+2 A \,b^{2} d \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}-6 B \,a^{2} e \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}+4 B a b d \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\right ) \sqrt {e x +d}}{3 \sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} \left (b x +a \right )^{\frac {3}{2}}}\)	\(503\)

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,method=_RETURNVERBOSE)

[Out]

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

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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)*(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (91) = 182\).
time = 1.38, size = 524, normalized size = 4.72 \begin {gather*} \left [\frac {\frac {3 \, {\left (B b^{3} d x^{2} + 2 \, B a b^{2} d x + B a^{2} b d - {\left (B a b^{2} x^{2} + 2 \, B a^{2} b x + B a^{3}\right )} e\right )} e^{\frac {1}{2}} \log \left (b^{2} d^{2} + \frac {4 \, {\left (b^{2} d + {\left (2 \, b^{2} x + a b\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} e^{\frac {1}{2}}}{\sqrt {b}} + {\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 )}{\sqrt {b}} - 4 \, {\left (3 \, B b^{2} d x + {\left (2 \, B a b + A b^{2}\right )} d - {\left (3 \, B a^{2} + {\left (4 \, B a b - A b^{2}\right )} x\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{6 \, {\left (b^{5} d x^{2} + 2 \, a b^{4} d x + a^{2} b^{3} d - {\left (a b^{4} x^{2} + 2 \, a^{2} b^{3} x + a^{3} b^{2}\right )} e\right )}}, -\frac {3 \, {\left (B b^{3} d x^{2} + 2 \, B a b^{2} d x + B a^{2} b d - {\left (B a b^{2} x^{2} + 2 \, B a^{2} b x + B a^{3}\right )} e\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {-\frac {e}{b}}}{2 \, {\left ({\left (b x^{2} + a x\right )} e^{2} + {\left (b d x + a d\right )} e\right )}}\right ) + 2 \, {\left (3 \, B b^{2} d x + {\left (2 \, B a b + A b^{2}\right )} d - {\left (3 \, B a^{2} + {\left (4 \, B a b - A b^{2}\right )} x\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{3 \, {\left (b^{5} d x^{2} + 2 \, a b^{4} d x + a^{2} b^{3} d - {\left (a b^{4} x^{2} + 2 \, a^{2} b^{3} x + a^{3} b^{2}\right )} e\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*b^3*d*x^2 + 2*B*a*b^2*d*x + B*a^2*b*d - (B*a*b^2*x^2 + 2*B*a^2*b*x + B*a^3)*e)*e^(1/2)*log(b^2*d^2
+ 4*(b^2*d + (2*b^2*x + a*b)*e)*sqrt(b*x + a)*sqrt(x*e + d)*e^(1/2)/sqrt(b) + (8*b^2*x^2 + 8*a*b*x + a^2)*e^2
+ 2*(4*b^2*d*x + 3*a*b*d)*e)/sqrt(b) - 4*(3*B*b^2*d*x + (2*B*a*b + A*b^2)*d - (3*B*a^2 + (4*B*a*b - A*b^2)*x)*
e)*sqrt(b*x + a)*sqrt(x*e + d))/(b^5*d*x^2 + 2*a*b^4*d*x + a^2*b^3*d - (a*b^4*x^2 + 2*a^2*b^3*x + a^3*b^2)*e),
-1/3*(3*(B*b^3*d*x^2 + 2*B*a*b^2*d*x + B*a^2*b*d - (B*a*b^2*x^2 + 2*B*a^2*b*x + B*a^3)*e)*sqrt(-e/b)*arctan(1
/2*(b*d + (2*b*x + a)*e)*sqrt(b*x + a)*sqrt(x*e + d)*sqrt(-e/b)/((b*x^2 + a*x)*e^2 + (b*d*x + a*d)*e)) + 2*(3*
B*b^2*d*x + (2*B*a*b + A*b^2)*d - (3*B*a^2 + (4*B*a*b - A*b^2)*x)*e)*sqrt(b*x + a)*sqrt(x*e + d))/(b^5*d*x^2 +
2*a*b^4*d*x + a^2*b^3*d - (a*b^4*x^2 + 2*a^2*b^3*x + a^3*b^2)*e)]

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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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 522 vs. \(2 (91) = 182\).
time = 0.64, 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)

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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)

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