3.17.77 \(\int (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1677]

Optimal. Leaf size=96 \[ -\frac {2 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)}+\frac {2 b (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^2 (a+b x)} \]

[Out]

-2/5*(-a*e+b*d)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)+2/7*b*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

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Rubi [A]
time = 0.03, 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 = {660, 45} \begin {gather*} \frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^2 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(-2*(b*d - a*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)) + (2*b*(d + e*x)^(7/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(7*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*} \int (d+e x)^{3/2} \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)^{3/2} \, 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)^{3/2}}{e}+\frac {b^2 (d+e x)^{5/2}}{e}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)}+\frac {2 b (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^2 (a+b x)}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 48, normalized size = 0.50 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} (d+e x)^{5/2} (-2 b d+7 a e+5 b e x)}{35 e^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(-2*b*d + 7*a*e + 5*b*e*x))/(35*e^2*(a + b*x))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 2.
time = 0.48, size = 33, normalized size = 0.34

method	result	size
default	\(\frac {2 \,\mathrm {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {5}{2}} \left (5 b e x +7 a e -2 b d \right )}{35 e^{2}}\)	\(33\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (5 b e x +7 a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{35 e^{2} \left (b x +a \right )}\)	\(43\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (5 b \,e^{3} x^{3}+7 a \,e^{3} x^{2}+8 b d \,e^{2} x^{2}+14 a d \,e^{2} x +b \,d^{2} e x +7 a \,d^{2} e -2 b \,d^{3}\right ) \sqrt {e x +d}}{35 \left (b x +a \right ) e^{2}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/35*csgn(b*x+a)*(e*x+d)^(5/2)*(5*b*e*x+7*a*e-2*b*d)/e^2

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Maxima [A]
time = 0.31, size = 67, normalized size = 0.70 \begin {gather*} \frac {2}{35} \, {\left (5 \, b x^{3} e^{3} - 2 \, b d^{3} + 7 \, a d^{2} e + {\left (8 \, b d e^{2} + 7 \, a e^{3}\right )} x^{2} + {\left (b d^{2} e + 14 \, a d e^{2}\right )} x\right )} \sqrt {x e + d} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

2/35*(5*b*x^3*e^3 - 2*b*d^3 + 7*a*d^2*e + (8*b*d*e^2 + 7*a*e^3)*x^2 + (b*d^2*e + 14*a*d*e^2)*x)*sqrt(x*e + d)*
e^(-2)

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Fricas [A]
time = 2.13, size = 70, normalized size = 0.73 \begin {gather*} -\frac {2}{35} \, {\left (2 \, b d^{3} - {\left (5 \, b x^{3} + 7 \, a x^{2}\right )} e^{3} - 2 \, {\left (4 \, b d x^{2} + 7 \, a d x\right )} e^{2} - {\left (b d^{2} x + 7 \, a d^{2}\right )} e\right )} \sqrt {x e + d} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

-2/35*(2*b*d^3 - (5*b*x^3 + 7*a*x^2)*e^3 - 2*(4*b*d*x^2 + 7*a*d*x)*e^2 - (b*d^2*x + 7*a*d^2)*e)*sqrt(x*e + d)*
e^(-2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right )^{\frac {3}{2}} \sqrt {\left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*((b*x+a)**2)**(1/2),x)

[Out]

Integral((d + e*x)**(3/2)*sqrt((a + b*x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (67) = 134\).
time = 1.31, size = 239, normalized size = 2.49 \begin {gather*} \frac {2}{105} \, {\left (35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} b d^{2} e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 14 \, {\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 e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 105 \, \sqrt {x e + d} a d^{2} \mathrm {sgn}\left (b x + a\right ) + 70 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a d \mathrm {sgn}\left (b x + a\right ) + 3 \, {\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 e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 7 \, {\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 \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/105*(35*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*b*d^2*e^(-1)*sgn(b*x + a) + 14*(3*(x*e + d)^(5/2) - 10*(x*e +
d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*b*d*e^(-1)*sgn(b*x + a) + 105*sqrt(x*e + d)*a*d^2*sgn(b*x + a) + 70*((x*e +
d)^(3/2) - 3*sqrt(x*e + d)*d)*a*d*sgn(b*x + a) + 3*(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*e^(-1)*sgn(b*x + a) + 7*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt
(x*e + d)*d^2)*a*sgn(b*x + a))*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(d + e*x)^(3/2),x)

[Out]

int(((a + b*x)^2)^(1/2)*(d + e*x)^(3/2), x)

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