3.9.80 \(\int \frac {(d+e x)^{5/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx\) [880]

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 = 0.03, 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 = {671, 663} \begin {gather*} -\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 {gather*}

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 663

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 671

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 = 0.08, size = 56, normalized size = 0.47 \begin {gather*} -\frac {2 \sqrt {c \left (d^2-e^2 x^2\right )} \left (43 d^2+14 d e x+3 e^2 x^2\right )}{15 c e \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.54, size = 51, normalized size = 0.43

method	result	size
default	\(-\frac {2 \sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (3 e^{2} x^{2}+14 d x e +43 d^{2}\right )}{15 \sqrt {e x +d}\, c e}\)	\(51\)
gosper	\(-\frac {2 \left (-e x +d \right ) \left (3 e^{2} x^{2}+14 d x e +43 d^{2}\right ) \sqrt {e x +d}}{15 e \sqrt {-x^{2} c \,e^{2}+c \,d^{2}}}\)	\(55\)
risch	\(-\frac {2 \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, \left (3 e^{2} x^{2}+14 d x e +43 d^{2}\right ) \left (-e x +d \right )}{15 \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}\, e \sqrt {-c \left (e x -d \right )}}\)	\(93\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 57, normalized size = 0.48 \begin {gather*} \frac {2 \, {\left (3 \, \sqrt {c} x^{3} e^{3} + 11 \, \sqrt {c} d x^{2} e^{2} + 29 \, \sqrt {c} d^{2} x e - 43 \, \sqrt {c} d^{3}\right )} e^{\left (-1\right )}}{15 \, \sqrt {-x e + d} c} \end {gather*}

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)*x^3*e^3 + 11*sqrt(c)*d*x^2*e^2 + 29*sqrt(c)*d^2*x*e - 43*sqrt(c)*d^3)*e^(-1)/(sqrt(-x*e + d)*c
)

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Fricas [A]
time = 2.65, size = 58, normalized size = 0.49 \begin {gather*} -\frac {2 \, \sqrt {-c x^{2} e^{2} + c d^{2}} {\left (3 \, x^{2} e^{2} + 14 \, d x e + 43 \, d^{2}\right )} \sqrt {x e + d}}{15 \, {\left (c x e^{2} + c d e\right )}} \end {gather*}

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

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

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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Giac [A]
time = 3.84, size = 103, normalized size = 0.87 \begin {gather*} \frac {2}{15} \, {\left (\frac {32 \, \sqrt {2} \sqrt {c d} d^{2}}{c} - \frac {60 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d} d^{2}}{c} + \frac {20 \, {\left (-{\left (x e + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c d - 3 \, {\left ({\left (x e + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (x e + d\right )} c + 2 \, c d}}{c^{3}}\right )} e^{\left (-1\right )} \end {gather*}

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]

2/15*(32*sqrt(2)*sqrt(c*d)*d^2/c - 60*sqrt(-(x*e + d)*c + 2*c*d)*d^2/c + (20*(-(x*e + d)*c + 2*c*d)^(3/2)*c*d
- 3*((x*e + d)*c - 2*c*d)^2*sqrt(-(x*e + d)*c + 2*c*d))/c^3)*e^(-1)

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Mupad [B]
time = 0.61, size = 79, normalized size = 0.66 \begin {gather*} -\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}} \end {gather*}

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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