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

Optimal. Leaf size=150 \[ -\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^{5/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {770, 21, 43} \begin {gather*} -\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{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)^2}{5 e^3 (a+b x) (d+e x)^{5/2}} \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)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)*(d + e*x)^(5/2)) + (4*b*(b*d - a*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^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x
)*Sqrt[d + e*x])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {(a+b x) \left (a b+b^2 x\right )}{(d+e x)^{7/2}} \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^2}{(d+e x)^{7/2}} \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^{7/2}}-\frac {2 b (b d-a e)}{e^2 (d+e x)^{5/2}}+\frac {b^2}{e^2 (d+e x)^{3/2}}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac {4 b (b d-a 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^2 \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 = 0.04, size = 79, normalized size = 0.53 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (3 a^2 e^2+2 a b e (2 d+5 e x)+b^2 \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]*(3*a^2*e^2 + 2*a*b*e*(2*d + 5*e*x) + b^2*(8*d^2 + 20*d*e*x + 15*e^2*x^2)))/(15*e^3*(a +
b*x)*(d + e*x)^(5/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 22.69, size = 100, normalized size = 0.67 \begin {gather*} -\frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (3 a^2 e^2+10 a b e (d+e x)-6 a b d e+3 b^2 d^2+15 b^2 (d+e x)^2-10 b^2 d (d+e x)\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^2*d^2 - 6*a*b*d*e + 3*a^2*e^2 - 10*b^2*d*(d + e*x) + 10*a*b*e*(d + e*x) + 1
5*b^2*(d + e*x)^2))/(15*e^2*(d + e*x)^(5/2)*(a*e + b*e*x))

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 95, normalized size = 0.63 \begin {gather*} -\frac {2 \, {\left (15 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} + 4 \, a b d e + 3 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b 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^2*e^2*x^2 + 8*b^2*d^2 + 4*a*b*d*e + 3*a^2*e^2 + 10*(2*b^2*d*e + 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)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 108, normalized size = 0.72 \begin {gather*} -\frac {2 \, {\left (15 \, {\left (x e + d\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, {\left (x e + d\right )} b^{2} d \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (x e + d\right )} a b e \mathrm {sgn}\left (b x + a\right ) - 6 \, a b d e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} 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^2*sgn(b*x + a) - 10*(x*e + d)*b^2*d*sgn(b*x + a) + 3*b^2*d^2*sgn(b*x + a) + 10*(x*e +
d)*a*b*e*sgn(b*x + a) - 6*a*b*d*e*sgn(b*x + a) + 3*a^2*e^2*sgn(b*x + a))*e^(-3)/(x*e + d)^(5/2)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 79, normalized size = 0.53 \begin {gather*} -\frac {2 \left (15 b^{2} x^{2} e^{2}+10 a b \,e^{2} x +20 b^{2} d e x +3 a^{2} e^{2}+4 a b d e +8 b^{2} 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^2*e^2*x^2+10*a*b*e^2*x+20*b^2*d*e*x+3*a^2*e^2+4*a*b*d*e+8*b^2*d^2)*((b*x+a)^2)^(1/2)
/e^3/(b*x+a)

________________________________________________________________________________________

maxima [A]  time = 0.61, size = 118, normalized size = 0.79 \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))

________________________________________________________________________________________

mupad [B]  time = 2.63, size = 151, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {4\,x\,\left (a\,e+2\,b\,d\right )}{3\,e^4}+\frac {2\,b\,x^2}{e^3}+\frac {\frac {2\,a^2\,e^2}{5}+\frac {8\,a\,b\,d\,e}{15}+\frac {16\,b^2\,d^2}{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 (a\,e^5+2\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{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)*((4*x*(a*e + 2*b*d))/(3*e^4) + (2*b*x^2)/e^3 + ((2*a^2*e^2)/5 + (16*b^2*d^2)/15 + (8*a*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*(a*e^5 + 2*b*d*e^4)*(d + e*x
)^(1/2))/(b*e^5) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))

________________________________________________________________________________________

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

________________________________________________________________________________________