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

Optimal. Leaf size=194 \[ \frac {2 (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {2 (a+b x) (B d-A e)}{3 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}-\frac {2 \sqrt {b} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 78, 51, 63, 208} \begin {gather*} \frac {2 (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {2 (a+b x) (B d-A e)}{3 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}-\frac {2 \sqrt {b} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 102, normalized size = 0.53 \begin {gather*} \frac {2 (a+b x) \left (3 e (d+e x) (A b-a B) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )-(b d-a e) (B d-A e)\right )}{3 e \sqrt {(a+b x)^2} (d+e x)^{3/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 40.29, size = 173, normalized size = 0.89 \begin {gather*} \frac {(-a e-b e x) \left (\frac {2 \left (A b^{3/2}-a \sqrt {b} B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{5/2}}+\frac {2 \left (a A e^2+3 a B e (d+e x)-a B d e-3 A b e (d+e x)-A b d e+b B d^2\right )}{3 e (d+e x)^{3/2} (a e-b d)^2}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 506, normalized size = 2.61 \begin {gather*} \left [-\frac {3 \, {\left ({\left (B a - A b\right )} e^{3} x^{2} + 2 \, {\left (B a - A b\right )} d e^{2} x + {\left (B a - A b\right )} d^{2} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) + 2 \, {\left (B b d^{2} + A a e^{2} + 3 \, {\left (B a - A b\right )} e^{2} x + 2 \, {\left (B a - 2 \, A b\right )} d e\right )} \sqrt {e x + d}}{3 \, {\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3} + {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x\right )}}, \frac {2 \, {\left (3 \, {\left ({\left (B a - A b\right )} e^{3} x^{2} + 2 \, {\left (B a - A b\right )} d e^{2} x + {\left (B a - A b\right )} d^{2} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (B b d^{2} + A a e^{2} + 3 \, {\left (B a - A b\right )} e^{2} x + 2 \, {\left (B a - 2 \, A b\right )} d e\right )} \sqrt {e x + d}\right )}}{3 \, {\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3} + {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 209, normalized size = 1.08 \begin {gather*} -\frac {2 \, {\left (B a b \mathrm {sgn}\left (b x + a\right ) - A b^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (B b d^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, {\left (x e + d\right )} B a e \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (x e + d\right )} A b e \mathrm {sgn}\left (b x + a\right ) - B a d e \mathrm {sgn}\left (b x + a\right ) - A b d e \mathrm {sgn}\left (b x + a\right ) + A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} {\left (x e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a*b*sgn(b*x + a) - A*b^2*sgn(b*x + a))*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^2*d^2 - 2*a*b*d*
e + a^2*e^2)*sqrt(-b^2*d + a*b*e)) - 2/3*(B*b*d^2*sgn(b*x + a) + 3*(x*e + d)*B*a*e*sgn(b*x + a) - 3*(x*e + d)*
A*b*e*sgn(b*x + a) - B*a*d*e*sgn(b*x + a) - A*b*d*e*sgn(b*x + a) + A*a*e^2*sgn(b*x + a))/((b^2*d^2*e - 2*a*b*d
*e^2 + a^2*e^3)*(x*e + d)^(3/2))

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maple [A]  time = 0.07, size = 235, normalized size = 1.21 \begin {gather*} \frac {2 \left (b x +a \right ) \left (3 \left (e x +d \right )^{\frac {3}{2}} A \,b^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 \sqrt {\left (a e -b d \right ) b}\, A b \,e^{2} x -3 \left (e x +d \right )^{\frac {3}{2}} B a b e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-3 \sqrt {\left (a e -b d \right ) b}\, B a \,e^{2} x -\sqrt {\left (a e -b d \right ) b}\, A a \,e^{2}+4 \sqrt {\left (a e -b d \right ) b}\, A b d e -2 \sqrt {\left (a e -b d \right ) b}\, B a d e -\sqrt {\left (a e -b d \right ) b}\, B b \,d^{2}\right )}{3 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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