3.19.69 \(\int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=152 \[ \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}{5 e^3 (a+b x)} \]

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Rubi [A]  time = 0.07, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {770, 21, 43} \begin {gather*} \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^3 (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)^2}{5 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)^2*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) - (4*b*(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*b^2*(d + e*x)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(9*e^3*(a + b*x))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, 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 770

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*} \int (a+b x) (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 (a+b x) \left (a b+b^2 x\right ) (d+e x)^{3/2} \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int (a+b x)^2 (d+e x)^{3/2} \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(-b d+a e)^2 (d+e x)^{3/2}}{e^2}-\frac {2 b (b d-a e) (d+e x)^{5/2}}{e^2}+\frac {b^2 (d+e x)^{7/2}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac {2 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}-\frac {4 b (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 b^2 (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 79, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} (d+e x)^{5/2} \left (63 a^2 e^2+18 a b e (5 e x-2 d)+b^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*(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)*(63*a^2*e^2 + 18*a*b*e*(-2*d + 5*e*x) + b^2*(8*d^2 - 20*d*e*x + 35*e^2*x^
2)))/(315*e^3*(a + b*x))

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IntegrateAlgebraic [A]  time = 33.67, size = 100, normalized size = 0.66 \begin {gather*} \frac {2 (d+e x)^{5/2} \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (63 a^2 e^2+90 a b e (d+e x)-126 a b d e+63 b^2 d^2+35 b^2 (d+e x)^2-90 b^2 d (d+e x)\right )}{315 e^2 (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(5/2)*Sqrt[(a*e + b*e*x)^2/e^2]*(63*b^2*d^2 - 126*a*b*d*e + 63*a^2*e^2 - 90*b^2*d*(d + e*x) + 90*
a*b*e*(d + e*x) + 35*b^2*(d + e*x)^2))/(315*e^2*(a*e + b*e*x))

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fricas [A]  time = 0.40, size = 137, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (35 \, b^{2} e^{4} x^{4} + 8 \, b^{2} d^{4} - 36 \, a b d^{3} e + 63 \, a^{2} d^{2} e^{2} + 10 \, {\left (5 \, b^{2} d e^{3} + 9 \, a b e^{4}\right )} x^{3} + 3 \, {\left (b^{2} d^{2} e^{2} + 48 \, a b d e^{3} + 21 \, a^{2} e^{4}\right )} x^{2} - 2 \, {\left (2 \, b^{2} d^{3} e - 9 \, a b d^{2} e^{2} - 63 \, a^{2} d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^2*e^4*x^4 + 8*b^2*d^4 - 36*a*b*d^3*e + 63*a^2*d^2*e^2 + 10*(5*b^2*d*e^3 + 9*a*b*e^4)*x^3 + 3*(b^2*
d^2*e^2 + 48*a*b*d*e^3 + 21*a^2*e^4)*x^2 - 2*(2*b^2*d^3*e - 9*a*b*d^2*e^2 - 63*a^2*d*e^3)*x)*sqrt(e*x + d)/e^3

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giac [B]  time = 0.21, size = 434, normalized size = 2.86 \begin {gather*} \frac {2}{315} \, {\left (210 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a b d^{2} e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 21 \, {\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^{2} d^{2} e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + 84 \, {\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 b d e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 18 \, {\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^{2} d e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + 315 \, \sqrt {x e + d} a^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 210 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a^{2} d \mathrm {sgn}\left (b x + a\right ) + 18 \, {\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 b e^{\left (-1\right )} \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^{2} e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + 21 \, {\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^{2} \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((b*x+a)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/315*(210*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a*b*d^2*e^(-1)*sgn(b*x + a) + 21*(3*(x*e + d)^(5/2) - 10*(x*e
+ d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*b^2*d^2*e^(-2)*sgn(b*x + a) + 84*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)
*d + 15*sqrt(x*e + d)*d^2)*a*b*d*e^(-1)*sgn(b*x + a) + 18*(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^2*d*e^(-2)*sgn(b*x + a) + 315*sqrt(x*e + d)*a^2*d^2*sgn(b*x + a) + 21
0*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a^2*d*sgn(b*x + a) + 18*(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*b*e^(-1)*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^2*e^(-2)*sgn(b*x + a) + 21
*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a^2*sgn(b*x + a))*e^(-1)

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maple [A]  time = 0.05, size = 79, normalized size = 0.52 \begin {gather*} \frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 b^{2} x^{2} e^{2}+90 a b \,e^{2} x -20 b^{2} d e x +63 a^{2} e^{2}-36 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{315 \left (b x +a \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(e*x+d)^(5/2)*(35*b^2*e^2*x^2+90*a*b*e^2*x-20*b^2*d*e*x+63*a^2*e^2-36*a*b*d*e+8*b^2*d^2)*((b*x+a)^2)^(1/
2)/e^3/(b*x+a)

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maxima [A]  time = 0.62, size = 167, normalized size = 1.10 \begin {gather*} \frac {2 \, {\left (5 \, b e^{3} x^{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 {e x + d} a}{35 \, e^{2}} + \frac {2 \, {\left (35 \, b e^{4} x^{4} + 8 \, b d^{4} - 18 \, a d^{3} e + 5 \, {\left (10 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (b d^{2} e^{2} + 24 \, a d e^{3}\right )} x^{2} - {\left (4 \, b d^{3} e - 9 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d} b}{315 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b*e^3*x^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(e*x + d)*
a/e^2 + 2/315*(35*b*e^4*x^4 + 8*b*d^4 - 18*a*d^3*e + 5*(10*b*d*e^3 + 9*a*e^4)*x^3 + 3*(b*d^2*e^2 + 24*a*d*e^3)
*x^2 - (4*b*d^3*e - 9*a*d^2*e^2)*x)*sqrt(e*x + d)*b/e^3

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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 (a+b\,x\right )\,{\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)*(a + b*x)*(d + e*x)^(3/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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