3.23.5 \(\int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx\) [2205]

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

[Out]

-2/3*(-A*e+B*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)/(e*x+d)^(3/2)+2*B*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/
2))*b^(1/2)/e^(5/2)-2*B*(b*x+a)^(1/2)/e^2/(e*x+d)^(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 (a+b x)^{3/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}+\frac {2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}}-\frac {2 B \sqrt {a+b x}}{e^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.21, size = 116, normalized size = 1.05 \begin {gather*} -\frac {2 \sqrt {a+b x} \left (A b e^2 x-b B d (3 d+4 e x)+a e (2 B d+A e+3 B e x)\right )}{3 e^2 (-b d+a e) (d+e x)^{3/2}}+\frac {2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x]*(A*b*e^2*x - b*B*d*(3*d + 4*e*x) + a*e*(2*B*d + A*e + 3*B*e*x)))/(3*e^2*(-(b*d) + a*e)*(d +
e*x)^(3/2)) + (2*Sqrt[b]*B*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(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 = 0.09, size = 503, normalized size = 4.53

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x+a)^(1/2)/(e*x+d)^(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*e^3*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^2*d*e^2*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*b*d*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))*b^2*d^2*e*x+2*A*b*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(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*b*d^2*e+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^2*d^3+6*B*a*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-8*B*b*d*e
*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+2*A*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+4*B*a*d*e*((b*x+a)*(e*x+d
))^(1/2)*(b*e)^(1/2)-6*B*b*d^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))*(b*x+a)^(1/2)/(b*e)^(1/2)/(a*e-b*d)/((b*x+
a)*(e*x+d))^(1/2)/e^2/(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)*(b*x+a)^(1/2)/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (90) = 180\).
time = 3.50, size = 494, normalized size = 4.45 \begin {gather*} \left [\frac {3 \, {\left (B b d^{3} - B a x^{2} e^{3} + {\left (B b d x^{2} - 2 \, B a d x\right )} e^{2} + {\left (2 \, B b d^{2} x - B a 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 (3 \, B b d^{2} - {\left (A a + {\left (3 \, B a + A b\right )} x\right )} e^{2} + 2 \, {\left (2 \, B b d x - B a d\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{6 \, {\left (b d^{3} e^{2} - a x^{2} e^{5} + {\left (b d x^{2} - 2 \, a d x\right )} e^{4} + {\left (2 \, b d^{2} x - a d^{2}\right )} e^{3}\right )}}, -\frac {3 \, {\left (B b d^{3} - B a x^{2} e^{3} + {\left (B b d x^{2} - 2 \, B a d x\right )} e^{2} + {\left (2 \, B b d^{2} x - B a 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 (3 \, B b d^{2} - {\left (A a + {\left (3 \, B a + A b\right )} x\right )} e^{2} + 2 \, {\left (2 \, B b d x - B a d\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{3 \, {\left (b d^{3} e^{2} - a x^{2} e^{5} + {\left (b d x^{2} - 2 \, a d x\right )} e^{4} + {\left (2 \, b d^{2} x - a d^{2}\right )} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 {a + b x}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (90) = 180\).
time = 1.01, size = 198, normalized size = 1.78 \begin {gather*} -\frac {2 \, B {\left | b \right |} e^{\left (-\frac {5}{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 {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, B b^{4} d {\left | b \right |} e^{2} - 3 \, B a b^{3} {\left | b \right |} e^{3} - A b^{4} {\left | b \right |} e^{3}\right )} {\left (b x + a\right )}}{b^{3} d e^{3} - a b^{2} e^{4}} + \frac {3 \, {\left (B b^{5} d^{2} {\left | b \right |} e - 2 \, B a b^{4} d {\left | b \right |} e^{2} + B a^{2} b^{3} {\left | b \right |} e^{3}\right )}}{b^{3} d e^{3} - a b^{2} e^{4}}\right )}}{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)*(b*x+a)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

-2*B*abs(b)*e^(-5/2)*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) -
2/3*sqrt(b*x + a)*((4*B*b^4*d*abs(b)*e^2 - 3*B*a*b^3*abs(b)*e^3 - A*b^4*abs(b)*e^3)*(b*x + a)/(b^3*d*e^3 - a*b
^2*e^4) + 3*(B*b^5*d^2*abs(b)*e - 2*B*a*b^4*d*abs(b)*e^2 + B*a^2*b^3*abs(b)*e^3)/(b^3*d*e^3 - a*b^2*e^4))/(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 = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {a+b\,x}}{{\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)^(1/2))/(d + e*x)^(5/2),x)

[Out]

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

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