\(\int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1678]

Optimal result

Integrand size = 30, antiderivative size = 96 \[ \int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {2 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x)}+\frac {2 b (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)} \]

[Out]

-2/3*(-a*e+b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)+2/5*b*(e*x+d)^(5/2)*((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} \[ \int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^2 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^2 (a+b x)} \]

[In]

Int[Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(-2*(b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^2*(a + b*x)) + (2*b*(d + e*x)^(5/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(5*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 ) \sqrt {d+e x} \, 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) \sqrt {d+e x}}{e}+\frac {b^2 (d+e x)^{3/2}}{e}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {2 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x)}+\frac {2 b (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.50 \[ \int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{3/2} (-2 b d+5 a e+3 b e x)}{15 e^2 (a+b x)} \]

[In]

Integrate[Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(3/2)*(-2*b*d + 5*a*e + 3*b*e*x))/(15*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 = 2.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.34

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {3}{2}} \left (3 b e x +5 a e -2 b d \right )}{15 e^{2}}\)	\(33\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (3 b e x +5 a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{2} \left (b x +a \right )}\)	\(43\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (3 b \,e^{2} x^{2}+5 a \,e^{2} x +b d e x +5 a d e -2 b \,d^{2}\right ) \sqrt {e x +d}}{15 \left (b x +a \right ) e^{2}}\)	\(62\)

[In]

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

[Out]

2/15*csgn(b*x+a)*(e*x+d)^(3/2)*(3*b*e*x+5*a*e-2*b*d)/e^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.48 \[ \int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e + {\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{2}} \]

[In]

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

[Out]

2/15*(3*b*e^2*x^2 - 2*b*d^2 + 5*a*d*e + (b*d*e + 5*a*e^2)*x)*sqrt(e*x + d)/e^2

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(d + e*x)*sqrt((a + b*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.48 \[ \int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e + {\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{2}} \]

[In]

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

[Out]

2/15*(3*b*e^2*x^2 - 2*b*d^2 + 5*a*d*e + (b*d*e + 5*a*e^2)*x)*sqrt(e*x + d)/e^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.29 \[ \int \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a d \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a \mathrm {sgn}\left (b x + a\right ) + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b \mathrm {sgn}\left (b x + a\right )}{e}\right )}}{15 \, e} \]

[In]

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

[Out]

2/15*(15*sqrt(e*x + d)*a*d*sgn(b*x + a) + 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*sgn(b*x + a) + 5*((e*x + d
)^(3/2) - 3*sqrt(e*x + d)*d)*b*d*sgn(b*x + a)/e + (3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)
*d^2)*b*sgn(b*x + a)/e)/e

Mupad [F(-1)]

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

[In]

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

[Out]

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