3.1842 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=162 \[ \frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^{5/2}}-\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.0767258, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {770, 77} \[ \frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^{5/2}}-\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(7/2),x]

[Out]

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

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]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right ) (A+B x)}{(d+e x)^{7/2}} \, dx}{a b+b^2 x}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (-\frac{b (b d-a e) (-B d+A e)}{e^2 (d+e x)^{7/2}}+\frac{b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^{5/2}}+\frac{b^2 B}{e^2 (d+e x)^{3/2}}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 (b d-a e) (B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac{2 (2 b B d-A b e-a B e) \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.0660683, size = 86, normalized size = 0.53 \[ -\frac{2 \sqrt{(a+b x)^2} \left (a e (3 A e+2 B d+5 B e x)+A b e (2 d+5 e x)+b B \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (a+b x) (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(A*b*e*(2*d + 5*e*x) + a*e*(2*B*d + 3*A*e + 5*B*e*x) + b*B*(8*d^2 + 20*d*e*x + 15*e^2*x^
2)))/(15*e^3*(a + b*x)*(d + e*x)^(5/2))

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Maple [A]  time = 0.004, size = 89, normalized size = 0.6 \begin{align*} -{\frac{30\,B{x}^{2}b{e}^{2}+10\,Axb{e}^{2}+10\,aB{e}^{2}x+40\,Bxbde+6\,aA{e}^{2}+4\,Abde+4\,aBde+16\,Bb{d}^{2}}{15\, \left ( bx+a \right ){e}^{3}}\sqrt{ \left ( bx+a \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x)

[Out]

-2/15/(e*x+d)^(5/2)*(15*B*b*e^2*x^2+5*A*b*e^2*x+5*B*a*e^2*x+20*B*b*d*e*x+3*A*a*e^2+2*A*b*d*e+2*B*a*d*e+8*B*b*d
^2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

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Maxima [A]  time = 1.04994, size = 159, normalized size = 0.98 \begin{align*} -\frac{2 \,{\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} A}{15 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt{e x + d}} - \frac{2 \,{\left (15 \, b e^{2} x^{2} + 8 \, b d^{2} + 2 \, a d e + 5 \,{\left (4 \, b d e + a e^{2}\right )} x\right )} B}{15 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )} \sqrt{e x + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

-2/15*(5*b*e*x + 2*b*d + 3*a*e)*A/((e^4*x^2 + 2*d*e^3*x + d^2*e^2)*sqrt(e*x + d)) - 2/15*(15*b*e^2*x^2 + 8*b*d
^2 + 2*a*d*e + 5*(4*b*d*e + a*e^2)*x)*B/((e^5*x^2 + 2*d*e^4*x + d^2*e^3)*sqrt(e*x + d))

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Fricas [A]  time = 1.26738, size = 224, normalized size = 1.38 \begin{align*} -\frac{2 \,{\left (15 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 3 \, A a e^{2} + 2 \,{\left (B a + A b\right )} d e + 5 \,{\left (4 \, B b d e +{\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(15*B*b*e^2*x^2 + 8*B*b*d^2 + 3*A*a*e^2 + 2*(B*a + A*b)*d*e + 5*(4*B*b*d*e + (B*a + A*b)*e^2)*x)*sqrt(e*
x + d)/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.10924, size = 182, normalized size = 1.12 \begin{align*} -\frac{2 \,{\left (15 \,{\left (x e + d\right )}^{2} B b \mathrm{sgn}\left (b x + a\right ) - 10 \,{\left (x e + d\right )} B b d \mathrm{sgn}\left (b x + a\right ) + 3 \, B b d^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )} B a e \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )} A b e \mathrm{sgn}\left (b x + a\right ) - 3 \, B a d e \mathrm{sgn}\left (b x + a\right ) - 3 \, A b d e \mathrm{sgn}\left (b x + a\right ) + 3 \, A a e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{15 \,{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2/15*(15*(x*e + d)^2*B*b*sgn(b*x + a) - 10*(x*e + d)*B*b*d*sgn(b*x + a) + 3*B*b*d^2*sgn(b*x + a) + 5*(x*e + d
)*B*a*e*sgn(b*x + a) + 5*(x*e + d)*A*b*e*sgn(b*x + a) - 3*B*a*d*e*sgn(b*x + a) - 3*A*b*d*e*sgn(b*x + a) + 3*A*
a*e^2*sgn(b*x + a))*e^(-3)/(x*e + d)^(5/2)