3.18.37 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{11/2}} \, dx\)

Optimal. Leaf size=236 \[ -\frac {15 c^2 d^2 \sqrt {c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{4 e^{7/2}}+\frac {15 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^3 \sqrt {d+e x}}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {662, 664, 660, 205} \begin {gather*} \frac {15 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^3 \sqrt {d+e x}}-\frac {15 c^2 d^2 \sqrt {c d^2-a e^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 )}{4 e^{7/2}}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}} \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)^(11/2),x]

[Out]

(15*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e^3*Sqrt[d + e*x]) - (5*c*d*(a*d*e + (c*d^2 + a*e^
2)*x + c*d*e*x^2)^(3/2))/(4*e^2*(d + e*x)^(5/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(2*e*(d + e*x
)^(9/2)) - (15*c^2*d^2*Sqrt[c*d^2 - a*e^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])])/(4*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 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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)^{11/2}} \, dx &=-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}+\frac {(5 c d) \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)^{7/2}} \, dx}{4 e}\\ &=-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}+\frac {\left (15 c^2 d^2\right ) \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}{8 e^2}\\ &=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac {\left (15 c^2 d^2 \left (c d^2-a e^2\right )\right ) \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}{8 e^3}\\ &=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac {\left (15 c^2 d^2 \left (c d^2-a e^2\right )\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 )}{4 e^2}\\ &=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac {15 c^2 d^2 \sqrt {c d^2-a e^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 )}{4 e^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 83, normalized size = 0.35 \begin {gather*} \frac {2 c^2 d^2 ((d+e x) (a e+c d x))^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{7 (d+e x)^{7/2} \left (c d^2-a e^2\right )^3} \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)^(11/2),x]

[Out]

(2*c^2*d^2*((a*e + c*d*x)*(d + e*x))^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)
])/(7*(c*d^2 - a*e^2)^3*(d + e*x)^(7/2))

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IntegrateAlgebraic [F]  time = 180.03, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(11/2),x]

[Out]

$Aborted

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fricas [A]  time = 0.47, size = 562, normalized size = 2.38 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\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 (8 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + {\left (25 \, c^{2} d^{3} e - 9 \, 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}}{8 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}, \frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\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 (8 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + {\left (25 \, c^{2} d^{3} e - 9 \, 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}}{4 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} 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)^(11/2),x, algorithm="fricas")

[Out]

[1/8*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*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*(8*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 5*a*c*d^2*e^2 - 2*a^2*e^4 +
(25*c^2*d^3*e - 9*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^6*x^3 + 3*d*e^5
*x^2 + 3*d^2*e^4*x + d^3*e^3), 1/4*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*sqrt((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)) + (
8*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 5*a*c*d^2*e^2 - 2*a^2*e^4 + (25*c^2*d^3*e - 9*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^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)]

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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)^(11/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.08, size = 527, normalized size = 2.23 \begin {gather*} -\frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (15 a \,c^{2} d^{2} e^{4} x^{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^{4} e^{2} x^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+30 a \,c^{2} d^{3} e^{3} x \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-30 c^{3} d^{5} e x \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+15 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 )-8 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{2} e^{2} x^{2}+9 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a c d \,e^{3} x -25 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{3} e x +2 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a^{2} e^{4}+5 \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 )}{4 \left (e x +d \right )^{\frac {5}{2}} \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)^(11/2),x)

[Out]

-1/4*(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)*x^2*a*c^
2*d^2*e^4-15*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*x^2*c^3*d^4*e^2+30*arctanh((c*d*x+a*e)^(1/2)
/((a*e^2-c*d^2)*e)^(1/2)*e)*x*a*c^2*d^3*e^3-30*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*x*c^3*d^5*
e+15*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^6-8*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^2*e^2*x^2+9*(c*d*x+a*e)^(1/2)*((a*
e^2-c*d^2)*e)^(1/2)*a*c*d*e^3*x-25*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^3*e*x+2*((a*e^2-c*d^2)*e)^(
1/2)*(c*d*x+a*e)^(1/2)*a^2*e^4+5*((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)^(5/2)/(c*d*x+a*e)^(1/2)/e^3/((a*e^2-c*d^2)*e)^(1/2)

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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 {11}{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)^(11/2),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)^(11/2), x)

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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 )}^{11/2}} \,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)^(11/2),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(11/2), x)

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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)**(11/2),x)

[Out]

Timed out

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