3.2205 \(\int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx\)

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

[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 = 0.06, 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} \[ -\frac {2 (a+b x)^{3/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}-\frac {2 B \sqrt {a+b x}}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}} \]

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 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 {\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) \operatorname {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) \operatorname {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 = 1.03, size = 148, normalized size = 1.33 \[ \frac {2 \left (\frac {\sqrt {e} \sqrt {a+b x} \left (a e (A e+2 B d+3 B e x)+A b e^2 x-b B d (3 d+4 e x)\right )}{b d-a e}+\frac {3 B (b d-a e)^{3/2} \left (\frac {b (d+e x)}{b d-a e}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{b}\right )}{3 e^{5/2} (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 3.17, size = 515, normalized size = 4.64 \[ \left [\frac {3 \, {\left (B b d^{3} - B a d^{2} e + {\left (B b d e^{2} - B a e^{3}\right )} x^{2} + 2 \, {\left (B b d^{2} e - B a d e^{2}\right )} x\right )} \sqrt {\frac {b}{e}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e^{2} x + b d e + a e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {b}{e}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (3 \, B b d^{2} - 2 \, B a d e - A a e^{2} + {\left (4 \, B b d e - {\left (3 \, B a + A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (b d^{3} e^{2} - a d^{2} e^{3} + {\left (b d e^{4} - a e^{5}\right )} x^{2} + 2 \, {\left (b d^{2} e^{3} - a d e^{4}\right )} x\right )}}, -\frac {3 \, {\left (B b d^{3} - B a d^{2} e + {\left (B b d e^{2} - B a e^{3}\right )} x^{2} + 2 \, {\left (B b d^{2} e - B a d e^{2}\right )} x\right )} \sqrt {-\frac {b}{e}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {b}{e}}}{2 \, {\left (b^{2} e x^{2} + a b d + {\left (b^{2} d + a b e\right )} x\right )}}\right ) + 2 \, {\left (3 \, B b d^{2} - 2 \, B a d e - A a e^{2} + {\left (4 \, B b d e - {\left (3 \, B a + A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (b d^{3} e^{2} - a d^{2} e^{3} + {\left (b d e^{4} - a e^{5}\right )} x^{2} + 2 \, {\left (b d^{2} e^{3} - a d e^{4}\right )} x\right )}}\right ] \]

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

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giac [B]  time = 1.81, size = 198, normalized size = 1.78 \[ -\frac {2 \, B {\left | b \right |} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, B b^{4} d {\left | b \right |} e^{2} - 3 \, B a b^{3} {\left | b \right |} e^{3} - A b^{4} {\left | b \right |} e^{3}\right )} {\left (b x + a\right )}}{b^{3} d e^{3} - a b^{2} e^{4}} + \frac {3 \, {\left (B b^{5} d^{2} {\left | b \right |} e - 2 \, B a b^{4} d {\left | b \right |} e^{2} + B a^{2} b^{3} {\left | b \right |} e^{3}\right )}}{b^{3} d e^{3} - a b^{2} e^{4}}\right )}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \]

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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maple [B]  time = 0.03, size = 503, normalized size = 4.53 \[ -\frac {\left (-3 B a b \,e^{3} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+3 B \,b^{2} d \,e^{2} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-6 B a b d \,e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+6 B \,b^{2} d^{2} e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-3 B a b \,d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+3 B \,b^{2} d^{3} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b \,e^{2} x +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a \,e^{2} x -8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b d e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A a \,e^{2}+4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a d e -6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b \,d^{2}\right ) \sqrt {b x +a}}{3 \sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \left (e x +d \right )^{\frac {3}{2}} e^{2}} \]

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)

[Out]

-1/3*(-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^2*a*b*e^3+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^2*b^2*d*e^2-6*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*b*d*e^2+6*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*b^2*d^2*e+2*A*x*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-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))*a*b*d^2*e+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))*b^2*d^3+6*B*x*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-8*B*x*b*d
*e*((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 \[ \text {Exception raised: ValueError} \]

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(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 = -0.01 \[ \int \frac {\left (A+B\,x\right )\,\sqrt {a+b\,x}}{{\left (d+e\,x\right )}^{5/2}} \,d x \]

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

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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