3.880 \(\int \frac {(d+e x)^{5/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {64 d^2 \sqrt {c d^2-c e^2 x^2}}{15 c e \sqrt {d+e x}}-\frac {16 d \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{15 c e}-\frac {2 (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{5 c e} \]

[Out]

-2/5*(e*x+d)^(3/2)*(-c*e^2*x^2+c*d^2)^(1/2)/c/e-64/15*d^2*(-c*e^2*x^2+c*d^2)^(1/2)/c/e/(e*x+d)^(1/2)-16/15*d*(
e*x+d)^(1/2)*(-c*e^2*x^2+c*d^2)^(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} \[ -\frac {64 d^2 \sqrt {c d^2-c e^2 x^2}}{15 c e \sqrt {d+e x}}-\frac {16 d \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{15 c e}-\frac {2 (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{5 c e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/Sqrt[c*d^2 - c*e^2*x^2],x]

[Out]

(-64*d^2*Sqrt[c*d^2 - c*e^2*x^2])/(15*c*e*Sqrt[d + e*x]) - (16*d*Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2])/(15*c*
e) - (2*(d + e*x)^(3/2)*Sqrt[c*d^2 - c*e^2*x^2])/(5*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 \frac {(d+e x)^{5/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx &=-\frac {2 (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{5 c e}+\frac {1}{5} (8 d) \int \frac {(d+e x)^{3/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx\\ &=-\frac {16 d \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{15 c e}-\frac {2 (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{5 c e}+\frac {1}{15} \left (32 d^2\right ) \int \frac {\sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}} \, dx\\ &=-\frac {64 d^2 \sqrt {c d^2-c e^2 x^2}}{15 c e \sqrt {d+e x}}-\frac {16 d \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{15 c e}-\frac {2 (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{5 c e}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 59, normalized size = 0.50 \[ -\frac {2 (d-e x) \sqrt {d+e x} \left (43 d^2+14 d e x+3 e^2 x^2\right )}{15 e \sqrt {c \left (d^2-e^2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(5/2)/Sqrt[c*d^2 - c*e^2*x^2],x]

[Out]

(-2*(d - e*x)*Sqrt[d + e*x]*(43*d^2 + 14*d*e*x + 3*e^2*x^2))/(15*e*Sqrt[c*(d^2 - e^2*x^2)])

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fricas [A]  time = 0.75, size = 58, normalized size = 0.49 \[ -\frac {2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, e^{2} x^{2} + 14 \, d e x + 43 \, d^{2}\right )} \sqrt {e x + d}}{15 \, {\left (c e^{2} x + c d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

-2/15*sqrt(-c*e^2*x^2 + c*d^2)*(3*e^2*x^2 + 14*d*e*x + 43*d^2)*sqrt(e*x + d)/(c*e^2*x + c*d*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {5}{2}}}{\sqrt {-c e^{2} x^{2} + c d^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^(5/2)/sqrt(-c*e^2*x^2 + c*d^2), x)

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maple [A]  time = 0.05, size = 55, normalized size = 0.46 \[ -\frac {2 \left (-e x +d \right ) \left (3 e^{2} x^{2}+14 d x e +43 d^{2}\right ) \sqrt {e x +d}}{15 \sqrt {-c \,e^{2} x^{2}+c \,d^{2}}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x)

[Out]

-2/15*(-e*x+d)*(3*e^2*x^2+14*d*e*x+43*d^2)*(e*x+d)^(1/2)/e/(-c*e^2*x^2+c*d^2)^(1/2)

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maxima [A]  time = 1.48, size = 58, normalized size = 0.49 \[ \frac {2 \, {\left (3 \, \sqrt {c} e^{3} x^{3} + 11 \, \sqrt {c} d e^{2} x^{2} + 29 \, \sqrt {c} d^{2} e x - 43 \, \sqrt {c} d^{3}\right )}}{15 \, \sqrt {-e x + d} c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*sqrt(c)*e^3*x^3 + 11*sqrt(c)*d*e^2*x^2 + 29*sqrt(c)*d^2*e*x - 43*sqrt(c)*d^3)/(sqrt(-e*x + d)*c*e)

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mupad [B]  time = 0.61, size = 79, normalized size = 0.66 \[ -\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {2\,x^2\,\sqrt {d+e\,x}}{5\,c}+\frac {86\,d^2\,\sqrt {d+e\,x}}{15\,c\,e^2}+\frac {28\,d\,x\,\sqrt {d+e\,x}}{15\,c\,e}\right )}{x+\frac {d}{e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/(c*d^2 - c*e^2*x^2)^(1/2),x)

[Out]

-((c*d^2 - c*e^2*x^2)^(1/2)*((2*x^2*(d + e*x)^(1/2))/(5*c) + (86*d^2*(d + e*x)^(1/2))/(15*c*e^2) + (28*d*x*(d
+ e*x)^(1/2))/(15*c*e)))/(x + d/e)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{\frac {5}{2}}}{\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)/(-c*e**2*x**2+c*d**2)**(1/2),x)

[Out]

Integral((d + e*x)**(5/2)/sqrt(-c*(-d + e*x)*(d + e*x)), x)

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