3.863 \(\int (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2} \, dx\)

Optimal. Leaf size=119 \[ -\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e (d+e x)^{3/2}}-\frac {16 d \left (c d^2-c e^2 x^2\right )^{3/2}}{35 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{7 c e} \]

[Out]

-64/105*d^2*(-c*e^2*x^2+c*d^2)^(3/2)/c/e/(e*x+d)^(3/2)-16/35*d*(-c*e^2*x^2+c*d^2)^(3/2)/c/e/(e*x+d)^(1/2)-2/7*
(-c*e^2*x^2+c*d^2)^(3/2)*(e*x+d)^(1/2)/c/e

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Rubi [A]  time = 0.05, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {657, 649} \[ -\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e (d+e x)^{3/2}}-\frac {16 d \left (c d^2-c e^2 x^2\right )^{3/2}}{35 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{7 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*d^2*(c*d^2 - c*e^2*x^2)^(3/2))/(105*c*e*(d + e*x)^(3/2)) - (16*d*(c*d^2 - c*e^2*x^2)^(3/2))/(35*c*e*Sqrt[
d + e*x]) - (2*Sqrt[d + e*x]*(c*d^2 - c*e^2*x^2)^(3/2))/(7*c*e)

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rubi steps

\begin {align*} \int (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2} \, dx &=-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{7 c e}+\frac {1}{7} (8 d) \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx\\ &=-\frac {16 d \left (c d^2-c e^2 x^2\right )^{3/2}}{35 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{7 c e}+\frac {1}{35} \left (32 d^2\right ) \int \frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}} \, dx\\ &=-\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e (d+e x)^{3/2}}-\frac {16 d \left (c d^2-c e^2 x^2\right )^{3/2}}{35 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{7 c e}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 64, normalized size = 0.54 \[ \frac {2 \left (-71 d^3+17 d^2 e x+39 d e^2 x^2+15 e^3 x^3\right ) \sqrt {c \left (d^2-e^2 x^2\right )}}{105 e \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c*(d^2 - e^2*x^2)]*(-71*d^3 + 17*d^2*e*x + 39*d*e^2*x^2 + 15*e^3*x^3))/(105*e*Sqrt[d + e*x])

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fricas [A]  time = 0.94, size = 67, normalized size = 0.56 \[ \frac {2 \, {\left (15 \, e^{3} x^{3} + 39 \, d e^{2} x^{2} + 17 \, d^{2} e x - 71 \, d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{105 \, {\left (e^{2} x + d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*e^3*x^3 + 39*d*e^2*x^2 + 17*d^2*e*x - 71*d^3)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)/(e^2*x + d*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (e x + d\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)*(e*x + d)^(3/2), x)

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maple [A]  time = 0.05, size = 55, normalized size = 0.46 \[ -\frac {2 \left (-e x +d \right ) \left (15 e^{2} x^{2}+54 d e x +71 d^{2}\right ) \sqrt {-c \,e^{2} x^{2}+c \,d^{2}}}{105 \sqrt {e x +d}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(-c*e^2*x^2+c*d^2)^(1/2),x)

[Out]

-2/105*(-e*x+d)*(15*e^2*x^2+54*d*e*x+71*d^2)*(-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(1/2)/e

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maxima [A]  time = 1.55, size = 68, normalized size = 0.57 \[ \frac {2 \, {\left (15 \, \sqrt {c} e^{3} x^{3} + 39 \, \sqrt {c} d e^{2} x^{2} + 17 \, \sqrt {c} d^{2} e x - 71 \, \sqrt {c} d^{3}\right )} {\left (e x + d\right )} \sqrt {-e x + d}}{105 \, {\left (e^{2} x + d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*sqrt(c)*e^3*x^3 + 39*sqrt(c)*d*e^2*x^2 + 17*sqrt(c)*d^2*e*x - 71*sqrt(c)*d^3)*(e*x + d)*sqrt(-e*x +
d)/(e^2*x + d*e)

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mupad [B]  time = 0.55, size = 85, normalized size = 0.71 \[ \frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {26\,d\,x^2\,\sqrt {d+e\,x}}{35}-\frac {142\,d^3\,\sqrt {d+e\,x}}{105\,e^2}+\frac {2\,e\,x^3\,\sqrt {d+e\,x}}{7}+\frac {34\,d^2\,x\,\sqrt {d+e\,x}}{105\,e}\right )}{x+\frac {d}{e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(3/2),x)

[Out]

((c*d^2 - c*e^2*x^2)^(1/2)*((26*d*x^2*(d + e*x)^(1/2))/35 - (142*d^3*(d + e*x)^(1/2))/(105*e^2) + (2*e*x^3*(d
+ e*x)^(1/2))/7 + (34*d^2*x*(d + e*x)^(1/2))/(105*e)))/(x + d/e)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- c \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(-c*e**2*x**2+c*d**2)**(1/2),x)

[Out]

Integral(sqrt(-c*(-d + e*x)*(d + e*x))*(d + e*x)**(3/2), x)

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