\(\int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1836]

Optimal result

Integrand size = 35, antiderivative size = 164 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)}+\frac {2 b B (d+e x)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 e^3 (a+b x)} \]

[Out]

2/7*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-2/9*(-A*b*e-B*a*e+2*B*b*d)*(e*x+d)^(9/2)
*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/11*b*B*(e*x+d)^(11/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {784, 78} \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (-a B e-A b e+2 b B d)}{9 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e) (B d-A e)}{7 e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{11/2}}{11 e^3 (a+b x)} \]

[In]

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

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(11/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(11*e^3*(a + b*x))

Rule 78

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 784

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*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right ) (A+B x) (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) (-B d+A e) (d+e x)^{5/2}}{e^2}+\frac {b (-2 b B d+A b e+a B e) (d+e x)^{7/2}}{e^2}+\frac {b^2 B (d+e x)^{9/2}}{e^2}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e) (B d-A e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)}+\frac {2 b B (d+e x)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 e^3 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{7/2} \left (11 A b e (-2 d+7 e x)+11 a e (-2 B d+9 A e+7 B e x)+b B \left (8 d^2-28 d e x+63 e^2 x^2\right )\right )}{693 e^3 (a+b x)} \]

[In]

Integrate[(A + B*x)*(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)*(11*A*b*e*(-2*d + 7*e*x) + 11*a*e*(-2*B*d + 9*A*e + 7*B*e*x) + b*B*(8*d^2
- 28*d*e*x + 63*e^2*x^2)))/(693*e^3*(a + b*x))

Maple [C] (warning: unable to verify)

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

Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.48

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {7}{2}} \left (63 B b \,e^{2} x^{2}+77 A b \,e^{2} x +77 B a \,e^{2} x -28 B b d e x +99 A a \,e^{2}-22 A b d e -22 B a d e +8 B b \,d^{2}\right )}{693 e^{3}}\)	\(79\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (63 B b \,e^{2} x^{2}+77 A b \,e^{2} x +77 B a \,e^{2} x -28 B b d e x +99 A a \,e^{2}-22 A b d e -22 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{693 e^{3} \left (b x +a \right )}\)	\(89\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (63 B b \,e^{5} x^{5}+77 A b \,e^{5} x^{4}+77 B a \,e^{5} x^{4}+161 B b d \,e^{4} x^{4}+99 A a \,e^{5} x^{3}+209 A b d \,e^{4} x^{3}+209 B a d \,e^{4} x^{3}+113 B b \,d^{2} e^{3} x^{3}+297 A a d \,e^{4} x^{2}+165 A b \,d^{2} e^{3} x^{2}+165 B a \,d^{2} e^{3} x^{2}+3 B b \,d^{3} e^{2} x^{2}+297 A a \,d^{2} e^{3} x +11 A b \,d^{3} e^{2} x +11 B a \,d^{3} e^{2} x -4 B b \,d^{4} e x +99 A a \,d^{3} e^{2}-22 A b \,d^{4} e -22 B a \,d^{4} e +8 B b \,d^{5}\right ) \sqrt {e x +d}}{693 \left (b x +a \right ) e^{3}}\)	\(241\)

[In]

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

[Out]

2/693*csgn(b*x+a)*(e*x+d)^(7/2)*(63*B*b*e^2*x^2+77*A*b*e^2*x+77*B*a*e^2*x-28*B*b*d*e*x+99*A*a*e^2-22*A*b*d*e-2
2*B*a*d*e+8*B*b*d^2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.15 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (63 \, B b e^{5} x^{5} + 8 \, B b d^{5} + 99 \, A a d^{3} e^{2} - 22 \, {\left (B a + A b\right )} d^{4} e + 7 \, {\left (23 \, B b d e^{4} + 11 \, {\left (B a + A b\right )} e^{5}\right )} x^{4} + {\left (113 \, B b d^{2} e^{3} + 99 \, A a e^{5} + 209 \, {\left (B a + A b\right )} d e^{4}\right )} x^{3} + 3 \, {\left (B b d^{3} e^{2} + 99 \, A a d e^{4} + 55 \, {\left (B a + A b\right )} d^{2} e^{3}\right )} x^{2} - {\left (4 \, B b d^{4} e - 297 \, A a d^{2} e^{3} - 11 \, {\left (B a + A b\right )} d^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{693 \, e^{3}} \]

[In]

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

[Out]

2/693*(63*B*b*e^5*x^5 + 8*B*b*d^5 + 99*A*a*d^3*e^2 - 22*(B*a + A*b)*d^4*e + 7*(23*B*b*d*e^4 + 11*(B*a + A*b)*e
^5)*x^4 + (113*B*b*d^2*e^3 + 99*A*a*e^5 + 209*(B*a + A*b)*d*e^4)*x^3 + 3*(B*b*d^3*e^2 + 99*A*a*d*e^4 + 55*(B*a
+ A*b)*d^2*e^3)*x^2 - (4*B*b*d^4*e - 297*A*a*d^2*e^3 - 11*(B*a + A*b)*d^3*e^2)*x)*sqrt(e*x + d)/e^3

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.30 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e + {\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} + {\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d} A}{63 \, e^{2}} + \frac {2 \, {\left (63 \, b e^{5} x^{5} + 8 \, b d^{5} - 22 \, a d^{4} e + 7 \, {\left (23 \, b d e^{4} + 11 \, a e^{5}\right )} x^{4} + {\left (113 \, b d^{2} e^{3} + 209 \, a d e^{4}\right )} x^{3} + 3 \, {\left (b d^{3} e^{2} + 55 \, a d^{2} e^{3}\right )} x^{2} - {\left (4 \, b d^{4} e - 11 \, a d^{3} e^{2}\right )} x\right )} \sqrt {e x + d} B}{693 \, e^{3}} \]

[In]

integrate((B*x+A)*(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)*A/e^2 + 2/693*(63*b*e^5*x^5 + 8*b*d^5 - 22*a*d^4*e + 7*(23*b*d*e^4 + 11*a
*e^5)*x^4 + (113*b*d^2*e^3 + 209*a*d*e^4)*x^3 + 3*(b*d^3*e^2 + 55*a*d^2*e^3)*x^2 - (4*b*d^4*e - 11*a*d^3*e^2)*
x)*sqrt(e*x + d)*B/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 831 vs. \(2 (119) = 238\).

Time = 0.29 (sec) , antiderivative size = 831, normalized size of antiderivative = 5.07 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

2/3465*(3465*sqrt(e*x + d)*A*a*d^3*sgn(b*x + a) + 3465*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*d^2*sgn(b*x +
a) + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d^3*sgn(b*x + a)/e + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x +
d)*d)*A*b*d^3*sgn(b*x + a)/e + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*a*d*sg
n(b*x + a) + 231*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*d^3*sgn(b*x + a)/e^2 +
693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*d^2*sgn(b*x + a)/e + 693*(3*(e*x + d
)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b*d^2*sgn(b*x + a)/e + 99*(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*a*sgn(b*x + a) + 297*(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*b*d^2*sgn(b*x + a)/e^2 + 297*(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*a*d*sgn(b*x + a)/e + 297*(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*b*d*sgn(b*x + a)/e +
33*(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*b*d*sgn(b*x + a)/e^2 + 11*(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*a*sgn(b*x + a)/e + 11*(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)*A*b*sgn(b*x + a)/e +
5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*
x + d)^(3/2)*d^4 - 693*sqrt(e*x + d)*d^5)*B*b*sgn(b*x + a)/e^2)/e

Mupad [F(-1)]

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

[In]

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

[Out]

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