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

Optimal. Leaf size=160 \[ -\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{231 c e (d+e x)^{3/2}}-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{33 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{5/2}}{11 c e}-\frac {256 d^3 \left (c d^2-c e^2 x^2\right )^{5/2}}{1155 c e (d+e x)^{5/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {657, 649} \begin {gather*} -\frac {256 d^3 \left (c d^2-c e^2 x^2\right )^{5/2}}{1155 c e (d+e x)^{5/2}}-\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{231 c e (d+e x)^{3/2}}-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{33 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{5/2}}{11 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*d^3*(c*d^2 - c*e^2*x^2)^(5/2))/(1155*c*e*(d + e*x)^(5/2)) - (64*d^2*(c*d^2 - c*e^2*x^2)^(5/2))/(231*c*e*
(d + e*x)^(3/2)) - (8*d*(c*d^2 - c*e^2*x^2)^(5/2))/(33*c*e*Sqrt[d + e*x]) - (2*Sqrt[d + e*x]*(c*d^2 - c*e^2*x^
2)^(5/2))/(11*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} \left (c d^2-c e^2 x^2\right )^{3/2} \, dx &=-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{5/2}}{11 c e}+\frac {1}{11} (12 d) \int \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2} \, dx\\ &=-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{33 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{5/2}}{11 c e}+\frac {1}{33} \left (32 d^2\right ) \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx\\ &=-\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{231 c e (d+e x)^{3/2}}-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{33 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{5/2}}{11 c e}+\frac {1}{231} \left (128 d^3\right ) \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx\\ &=-\frac {256 d^3 \left (c d^2-c e^2 x^2\right )^{5/2}}{1155 c e (d+e x)^{5/2}}-\frac {64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{231 c e (d+e x)^{3/2}}-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{33 c e \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{5/2}}{11 c e}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 73, normalized size = 0.46 \begin {gather*} -\frac {2 c (d-e x)^2 \left (533 d^3+755 d^2 e x+455 d e^2 x^2+105 e^3 x^3\right ) \sqrt {c \left (d^2-e^2 x^2\right )}}{1155 e \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(d - e*x)^2*Sqrt[c*(d^2 - e^2*x^2)]*(533*d^3 + 755*d^2*e*x + 455*d*e^2*x^2 + 105*e^3*x^3))/(1155*e*Sqrt[
d + e*x])

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IntegrateAlgebraic [A]  time = 0.53, size = 78, normalized size = 0.49 \begin {gather*} -\frac {2 \left (128 d^3+160 d^2 (d+e x)+140 d (d+e x)^2+105 (d+e x)^3\right ) \left (2 c d (d+e x)-c (d+e x)^2\right )^{5/2}}{1155 c e (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*c*d*(d + e*x) - c*(d + e*x)^2)^(5/2)*(128*d^3 + 160*d^2*(d + e*x) + 140*d*(d + e*x)^2 + 105*(d + e*x)^3
))/(1155*c*e*(d + e*x)^(5/2))

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fricas [A]  time = 0.41, size = 95, normalized size = 0.59 \begin {gather*} -\frac {2 \, {\left (105 \, c e^{5} x^{5} + 245 \, c d e^{4} x^{4} - 50 \, c d^{2} e^{3} x^{3} - 522 \, c d^{3} e^{2} x^{2} - 311 \, c d^{4} e x + 533 \, c d^{5}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{1155 \, {\left (e^{2} x + d e\right )}} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

-2/1155*(105*c*e^5*x^5 + 245*c*d*e^4*x^4 - 50*c*d^2*e^3*x^3 - 522*c*d^3*e^2*x^2 - 311*c*d^4*e*x + 533*c*d^5)*s
qrt(-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 \begin {gather*} \int {\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{2}}\,{d x} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 66, normalized size = 0.41 \begin {gather*} -\frac {2 \left (-e x +d \right ) \left (105 e^{3} x^{3}+455 e^{2} x^{2} d +755 d^{2} x e +533 d^{3}\right ) \left (-c \,e^{2} x^{2}+c \,d^{2}\right )^{\frac {3}{2}}}{1155 \left (e x +d \right )^{\frac {3}{2}} e} \end {gather*}

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)^(3/2),x)

[Out]

-2/1155*(-e*x+d)*(105*e^3*x^3+455*d*e^2*x^2+755*d^2*e*x+533*d^3)*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(3/2)

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maxima [A]  time = 1.60, size = 96, normalized size = 0.60 \begin {gather*} -\frac {2 \, {\left (105 \, c^{\frac {3}{2}} e^{5} x^{5} + 245 \, c^{\frac {3}{2}} d e^{4} x^{4} - 50 \, c^{\frac {3}{2}} d^{2} e^{3} x^{3} - 522 \, c^{\frac {3}{2}} d^{3} e^{2} x^{2} - 311 \, c^{\frac {3}{2}} d^{4} e x + 533 \, c^{\frac {3}{2}} d^{5}\right )} {\left (e x + d\right )} \sqrt {-e x + d}}{1155 \, {\left (e^{2} x + d e\right )}} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

-2/1155*(105*c^(3/2)*e^5*x^5 + 245*c^(3/2)*d*e^4*x^4 - 50*c^(3/2)*d^2*e^3*x^3 - 522*c^(3/2)*d^3*e^2*x^2 - 311*
c^(3/2)*d^4*e*x + 533*c^(3/2)*d^5)*(e*x + d)*sqrt(-e*x + d)/(e^2*x + d*e)

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mupad [B]  time = 0.64, size = 128, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {1066\,c\,d^5\,\sqrt {d+e\,x}}{1155\,e^2}-\frac {348\,c\,d^3\,x^2\,\sqrt {d+e\,x}}{385}+\frac {2\,c\,e^3\,x^5\,\sqrt {d+e\,x}}{11}-\frac {20\,c\,d^2\,e\,x^3\,\sqrt {d+e\,x}}{231}-\frac {622\,c\,d^4\,x\,\sqrt {d+e\,x}}{1155\,e}+\frac {14\,c\,d\,e^2\,x^4\,\sqrt {d+e\,x}}{33}\right )}{x+\frac {d}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c*d^2 - c*e^2*x^2)^(1/2)*((1066*c*d^5*(d + e*x)^(1/2))/(1155*e^2) - (348*c*d^3*x^2*(d + e*x)^(1/2))/385 + (
2*c*e^3*x^5*(d + e*x)^(1/2))/11 - (20*c*d^2*e*x^3*(d + e*x)^(1/2))/231 - (622*c*d^4*x*(d + e*x)^(1/2))/(1155*e
) + (14*c*d*e^2*x^4*(d + e*x)^(1/2))/33))/(x + d/e)

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

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)**(3/2),x)

[Out]

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

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