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

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

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Rubi [A]  time = 0.08, antiderivative size = 162, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]))))

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) \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*}

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Mathematica [A]  time = 0.06, size = 86, normalized size = 0.53 \begin {gather*} -\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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 40.00, size = 112, normalized size = 0.69 \begin {gather*} -\frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (3 a A e^2+5 a B e (d+e x)-3 a B d e+5 A b e (d+e x)-3 A b d e+3 b B d^2-10 b B d (d+e x)+15 b B (d+e x)^2\right )}{15 e^2 (d+e x)^{5/2} (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 101, normalized size = 0.62 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.29, size = 135, normalized size = 0.83 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.05, size = 89, normalized size = 0.55 \begin {gather*} -\frac {2 \left (15 B b \,x^{2} e^{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}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} \left (b x +a \right ) e^{3}} \end {gather*}

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 = 0.58, size = 118, normalized size = 0.73 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 2.52, size = 174, normalized size = 1.07 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^2}{e^3}+\frac {6\,A\,a\,e^2+16\,B\,b\,d^2+4\,A\,b\,d\,e+4\,B\,a\,d\,e}{15\,b\,e^5}+\frac {x\,\left (10\,A\,b\,e^2+10\,B\,a\,e^2+40\,B\,b\,d\,e\right )}{15\,b\,e^5}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (15\,a\,e^5+30\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{15\,b\,e^5}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((a + b*x)^2)^(1/2)*((2*B*x^2)/e^3 + (6*A*a*e^2 + 16*B*b*d^2 + 4*A*b*d*e + 4*B*a*d*e)/(15*b*e^5) + (x*(10*A*
b*e^2 + 10*B*a*e^2 + 40*B*b*d*e))/(15*b*e^5)))/(x^3*(d + e*x)^(1/2) + (a*d^2*(d + e*x)^(1/2))/(b*e^2) + (x^2*(
15*a*e^5 + 30*b*d*e^4)*(d + e*x)^(1/2))/(15*b*e^5) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))

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

[Out]

Timed out

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