3.17.18 \(\int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=164 \[ -\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^3 (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {770, 77} \begin {gather*} -\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-a B e-A b e+2 b B d)}{5 e^3 (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) (B d-A e)}{3 e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*Sqrt[d + e*x]*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)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(7/2)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x))

Rule 77

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

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Mathematica [A]  time = 0.06, size = 88, normalized size = 0.54 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} (d+e x)^{3/2} \left (7 a e (5 A e-2 B d+3 B e x)+7 A b e (3 e x-2 d)+b B \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*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)*(7*A*b*e*(-2*d + 3*e*x) + 7*a*e*(-2*B*d + 5*A*e + 3*B*e*x) + b*B*(8*d^2 -
12*d*e*x + 15*e^2*x^2)))/(105*e^3*(a + b*x))

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IntegrateAlgebraic [A]  time = 36.58, size = 112, normalized size = 0.68 \begin {gather*} \frac {2 (d+e x)^{3/2} \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (35 a A e^2+21 a B e (d+e x)-35 a B d e+21 A b e (d+e x)-35 A b d e+35 b B d^2-42 b B d (d+e x)+15 b B (d+e x)^2\right )}{105 e^2 (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(3/2)*Sqrt[(a*e + b*e*x)^2/e^2]*(35*b*B*d^2 - 35*A*b*d*e - 35*a*B*d*e + 35*a*A*e^2 - 42*b*B*d*(d
+ e*x) + 21*A*b*e*(d + e*x) + 21*a*B*e*(d + e*x) + 15*b*B*(d + e*x)^2))/(105*e^2*(a*e + b*e*x))

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fricas [A]  time = 0.42, size = 108, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (15 \, B b e^{3} x^{3} + 8 \, B b d^{3} + 35 \, A a d e^{2} - 14 \, {\left (B a + A b\right )} d^{2} e + 3 \, {\left (B b d e^{2} + 7 \, {\left (B a + A b\right )} e^{3}\right )} x^{2} - {\left (4 \, B b d^{2} e - 35 \, A a e^{3} - 7 \, {\left (B a + A b\right )} d e^{2}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*b*e^3*x^3 + 8*B*b*d^3 + 35*A*a*d*e^2 - 14*(B*a + A*b)*d^2*e + 3*(B*b*d*e^2 + 7*(B*a + A*b)*e^3)*x^
2 - (4*B*b*d^2*e - 35*A*a*e^3 - 7*(B*a + A*b)*d*e^2)*x)*sqrt(e*x + d)/e^3

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giac [B]  time = 0.24, size = 322, normalized size = 1.96 \begin {gather*} \frac {2}{105} \, {\left (35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} B a d e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} A b d 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 )} B b d e^{\left (-2\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 )} B a 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 b e^{\left (-1\right )} \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 b e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + 105 \, \sqrt {x e + d} A a d \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} A 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((B*x+A)*(e*x+d)^(1/2)*((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*a*d*e^(-1)*sgn(b*x + a) + 35*((x*e + d)^(3/2) - 3*sqrt(x*e +
d)*d)*A*b*d*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)*B*b*d*e
^(-2)*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)*B*a*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*b*e^(-1)*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*b*e^(-2)*sgn(b*x + a) + 10
5*sqrt(x*e + d)*A*a*d*sgn(b*x + a) + 35*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*A*a*sgn(b*x + a))*e^(-1)

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maple [A]  time = 0.05, size = 89, normalized size = 0.54 \begin {gather*} \frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 B b \,x^{2} e^{2}+21 A b \,e^{2} x +21 B a \,e^{2} x -12 B b d e x +35 A a \,e^{2}-14 A b d e -14 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{105 \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)^(1/2)*((b*x+a)^2)^(1/2),x)

[Out]

2/105*(e*x+d)^(3/2)*(15*B*b*e^2*x^2+21*A*b*e^2*x+21*B*a*e^2*x-12*B*b*d*e*x+35*A*a*e^2-14*A*b*d*e-14*B*a*d*e+8*
B*b*d^2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

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maxima [A]  time = 0.60, size = 120, normalized size = 0.73 \begin {gather*} \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} A}{15 \, e^{2}} + \frac {2 \, {\left (15 \, b e^{3} x^{3} + 8 \, b d^{3} - 14 \, a d^{2} e + 3 \, {\left (b d e^{2} + 7 \, a e^{3}\right )} x^{2} - {\left (4 \, b d^{2} e - 7 \, a d e^{2}\right )} x\right )} \sqrt {e x + d} B}{105 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(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)*A/e^2 + 2/105*(15*b*e^3*x^3 + 8*b*d
^3 - 14*a*d^2*e + 3*(b*d*e^2 + 7*a*e^3)*x^2 - (4*b*d^2*e - 7*a*d*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 )\,\sqrt {d+e\,x} \,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)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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