3.18.35 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=240 \[ \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac {2 \left (a-\frac {c d^2}{e^2}\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 (d+e x)^{3/2}}-\frac {2 \left (c d^2-a e^2\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{e^{7/2}}+\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3 \sqrt {d+e x}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {664, 660, 205} \begin {gather*} \frac {2 \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3 \sqrt {d+e x}}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac {2 \left (a-\frac {c d^2}{e^2}\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 (d+e x)^{3/2}}-\frac {2 \left (c d^2-a e^2\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(2*(c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e^3*Sqrt[d + e*x]) + (2*(a - (c*d^2)/e^2)*(
a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(3*(d + e*x)^(3/2)) + (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)
^(5/2))/(5*e*(d + e*x)^(5/2)) - (2*(c*d^2 - a*e^2)^(5/2)*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*
e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/e^(7/2)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx &=\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac {\left (2 c d^2 e-e \left (c d^2+a e^2\right )\right ) \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx}{e^2}\\ &=\frac {2 \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 (d+e x)^{3/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac {\left (c d^2-a e^2\right )^2 \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{3/2}} \, dx}{e^2}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 \sqrt {d+e x}}+\frac {2 \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 (d+e x)^{3/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac {\left (c d^2-a e^2\right )^3 \int \frac {1}{\sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{e^3}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 \sqrt {d+e x}}+\frac {2 \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 (d+e x)^{3/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac {\left (2 \left (c d^2-a e^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}}\right )}{e^2}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 \sqrt {d+e x}}+\frac {2 \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 (d+e x)^{3/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac {2 \left (c d^2-a e^2\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{e^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.31, size = 152, normalized size = 0.63 \begin {gather*} \frac {((d+e x) (a e+c d x))^{5/2} \left (\frac {10 \left (a e^2-c d^2\right ) \left (4 a e^2+c d (e x-3 d)\right )}{3 e^2 (a e+c d x)^2}-\frac {10 \left (c d^2-a e^2\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{e^{5/2} (a e+c d x)^{5/2}}+2\right )}{5 e (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(((a*e + c*d*x)*(d + e*x))^(5/2)*(2 + (10*(-(c*d^2) + a*e^2)*(4*a*e^2 + c*d*(-3*d + e*x)))/(3*e^2*(a*e + c*d*x
)^2) - (10*(c*d^2 - a*e^2)^(5/2)*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]])/(e^(5/2)*(a*e + c*d*
x)^(5/2))))/(5*e*(d + e*x)^(5/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 2.79, size = 223, normalized size = 0.93 \begin {gather*} \frac {((d+e x) (a e+c d x))^{5/2} \left (\frac {2 \left (15 a^2 e^4 \sqrt {a e+c d x}+15 c^2 d^4 \sqrt {a e+c d x}-30 a c d^2 e^2 \sqrt {a e+c d x}-5 c d^2 e (a e+c d x)^{3/2}+5 a e^3 (a e+c d x)^{3/2}+3 e^2 (a e+c d x)^{5/2}\right )}{15 e^3}-\frac {2 \left (c d^2-a e^2\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{e^{7/2}}\right )}{(d+e x)^{5/2} (a e+c d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(((a*e + c*d*x)*(d + e*x))^(5/2)*((2*(15*c^2*d^4*Sqrt[a*e + c*d*x] - 30*a*c*d^2*e^2*Sqrt[a*e + c*d*x] + 15*a^2
*e^4*Sqrt[a*e + c*d*x] - 5*c*d^2*e*(a*e + c*d*x)^(3/2) + 5*a*e^3*(a*e + c*d*x)^(3/2) + 3*e^2*(a*e + c*d*x)^(5/
2)))/(15*e^3) - (2*(c*d^2 - a*e^2)^(5/2)*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]])/e^(7/2)))/((
a*e + c*d*x)^(5/2)*(d + e*x)^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 538, normalized size = 2.24 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{e}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} e \sqrt {-\frac {c d^{2} - a e^{2}}{e}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (3 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 35 \, a c d^{2} e^{2} + 23 \, a^{2} e^{4} - {\left (5 \, c^{2} d^{3} e - 11 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{15 \, {\left (e^{4} x + d e^{3}\right )}}, \frac {2 \, {\left (15 \, {\left (c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{e}} \arctan \left (\frac {\sqrt {e x + d} \sqrt {\frac {c d^{2} - a e^{2}}{e}}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}\right ) + {\left (3 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 35 \, a c d^{2} e^{2} + 23 \, a^{2} e^{4} - {\left (5 \, c^{2} d^{3} e - 11 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}\right )}}{15 \, {\left (e^{4} x + d e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*(c^2*d^5 - 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-(c*d^2 - a*e^2
)/e)*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*
x + d)*e*sqrt(-(c*d^2 - a*e^2)/e))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(3*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 35*a*c*d^2
*e^2 + 23*a^2*e^4 - (5*c^2*d^3*e - 11*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))
/(e^4*x + d*e^3), 2/15*(15*(c^2*d^5 - 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqr
t((c*d^2 - a*e^2)/e)*arctan(sqrt(e*x + d)*sqrt((c*d^2 - a*e^2)/e)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))
+ (3*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 35*a*c*d^2*e^2 + 23*a^2*e^4 - (5*c^2*d^3*e - 11*a*c*d*e^3)*x)*sqrt(c*d*e*
x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^4*x + d*e^3)]

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.07, size = 437, normalized size = 1.82 \begin {gather*} -\frac {2 \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (15 a^{3} e^{6} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-45 a^{2} c \,d^{2} e^{4} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+45 a \,c^{2} d^{4} e^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-15 c^{3} d^{6} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-3 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{2} e^{2} x^{2}-11 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a c d \,e^{3} x +5 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{3} e x -23 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a^{2} e^{4}+35 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a c \,d^{2} e^{2}-15 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, c^{2} d^{4}\right )}{15 \sqrt {e x +d}\, \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2)/(e*x+d)^(7/2),x)

[Out]

-2/15*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(15*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*a^3*e^6
-45*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*a^2*c*d^2*e^4+45*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*
d^2)*e)^(1/2)*e)*a*c^2*d^4*e^2-15*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*c^3*d^6-3*(c*d*x+a*e)^(
1/2)*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^2*e^2*x^2-11*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*a*c*d*e^3*x+5*(c*d*x
+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^3*e*x-23*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a^2*e^4+35*((a*e^
2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d^2*e^2-15*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^4)/(e*x+d)^
(1/2)/(c*d*x+a*e)^(1/2)/e^3/((a*e^2-c*d^2)*e)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/(e*x + d)^(7/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{{\left (d+e\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(7/2),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(7/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________