\(\int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1676]

Optimal result

Integrand size = 30, antiderivative size = 96 \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {2 (b d-a e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^2 (a+b x)}+\frac {2 b (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^2 (a+b x)} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {660, 45} \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^2 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^2 (a+b x)} \]

[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 45

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 660

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.50 \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{7/2} (-2 b d+9 a e+7 b e x)}{63 e^2 (a+b x)} \]

[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))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 2.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.34

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {7}{2}} \left (7 b e x +9 a e -2 b d \right )}{63 e^{2}}\)	\(33\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (7 b e x +9 a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{63 e^{2} \left (b x +a \right )}\)	\(43\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (7 b \,e^{4} x^{4}+9 a \,e^{4} x^{3}+19 b d \,e^{3} x^{3}+27 a d \,e^{3} x^{2}+15 b \,d^{2} e^{2} x^{2}+27 a \,d^{2} e^{2} x +b \,d^{3} e x +9 a \,d^{3} e -2 b \,d^{4}\right ) \sqrt {e x +d}}{63 \left (b x +a \right ) e^{2}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97 \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e + {\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} + {\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{63 \, e^{2}} \]

[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

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97 \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e + {\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} + {\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{63 \, e^{2}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (66) = 132\).

Time = 0.29 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.69 \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a d^{3} \mathrm {sgn}\left (b x + a\right ) + 315 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a d^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b d^{3} \mathrm {sgn}\left (b x + a\right )}{e} + 63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a d \mathrm {sgn}\left (b x + a\right ) + \frac {63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b d^{2} \mathrm {sgn}\left (b x + a\right )}{e} + 9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a \mathrm {sgn}\left (b x + a\right ) + \frac {27 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b \mathrm {sgn}\left (b x + a\right )}{e}\right )}}{315 \, e} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{5/2} \,d x \]

[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)