3.1676 \(\int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=96 \[ \frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^2 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^2 (a+b x)} \]

[Out]

-2/7*(-a*e+b*d)*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)+2/9*b*(e*x+d)^(9/2)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

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Rubi [A]  time = 0.04, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {646, 43} \[ \frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^2 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^2*(a + b*x)) + (2*b*(d + e*x)^(9/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(9*e^2*(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 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right ) (d+e x)^{5/2} \, 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) (d+e x)^{5/2}}{e}+\frac {b^2 (d+e x)^{7/2}}{e}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 (b d-a e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^2 (a+b x)}+\frac {2 b (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^2 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 0.50 \[ \frac {2 \sqrt {(a+b x)^2} (d+e x)^{7/2} (9 a e-2 b d+7 b e x)}{63 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(7/2)*(-2*b*d + 9*a*e + 7*b*e*x))/(63*e^2*(a + b*x))

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fricas [A]  time = 1.00, size = 93, normalized size = 0.97 \[ \frac {2 \, {\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e + {\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} + {\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{63 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*b*e^4*x^4 - 2*b*d^4 + 9*a*d^3*e + (19*b*d*e^3 + 9*a*e^4)*x^3 + 3*(5*b*d^2*e^2 + 9*a*d*e^3)*x^2 + (b*d^
3*e + 27*a*d^2*e^2)*x)*sqrt(e*x + d)/e^2

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giac [B]  time = 0.21, size = 373, normalized size = 3.89 \[ \frac {2}{315} \, {\left (105 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} b d^{3} e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 63 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} b d^{2} e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 315 \, \sqrt {x e + d} a d^{3} \mathrm {sgn}\left (b x + a\right ) + 315 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a d^{2} \mathrm {sgn}\left (b x + a\right ) + 27 \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} b d e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 63 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} a d \mathrm {sgn}\left (b x + a\right ) + {\left (35 \, {\left (x e + d\right )}^{\frac {9}{2}} - 180 \, {\left (x e + d\right )}^{\frac {7}{2}} d + 378 \, {\left (x e + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {x e + d} d^{4}\right )} b e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 9 \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} a \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(105*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*b*d^3*e^(-1)*sgn(b*x + a) + 63*(3*(x*e + d)^(5/2) - 10*(x*e +
d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*b*d^2*e^(-1)*sgn(b*x + a) + 315*sqrt(x*e + d)*a*d^3*sgn(b*x + a) + 315*((x
*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a*d^2*sgn(b*x + a) + 27*(5*(x*e + d)^(7/2) - 21*(x*e + d)^(5/2)*d + 35*(x*e
+ d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*b*d*e^(-1)*sgn(b*x + a) + 63*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d
+ 15*sqrt(x*e + d)*d^2)*a*d*sgn(b*x + a) + (35*(x*e + d)^(9/2) - 180*(x*e + d)^(7/2)*d + 378*(x*e + d)^(5/2)*
d^2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*e + d)*d^4)*b*e^(-1)*sgn(b*x + a) + 9*(5*(x*e + d)^(7/2) - 21*(x*e
+ d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*a*sgn(b*x + a))*e^(-1)

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maple [A]  time = 0.04, size = 43, normalized size = 0.45 \[ \frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (7 b e x +9 a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{63 \left (b x +a \right ) e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(e*x+d)^(7/2)*(7*b*e*x+9*a*e-2*b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

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maxima [A]  time = 1.13, size = 93, normalized size = 0.97 \[ \frac {2 \, {\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e + {\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} + {\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{63 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*b*e^4*x^4 - 2*b*d^4 + 9*a*d^3*e + (19*b*d*e^3 + 9*a*e^4)*x^3 + 3*(5*b*d^2*e^2 + 9*a*d*e^3)*x^2 + (b*d^
3*e + 27*a*d^2*e^2)*x)*sqrt(e*x + d)/e^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((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 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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