3.2010 \(\int \frac {(d+e x)^{11/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=178 \[ -\frac {7 e \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}}+\frac {7 e \sqrt {d+e x} \left (c d^2-a e^2\right )^2}{c^4 d^4}+\frac {7 e (d+e x)^{3/2} \left (c d^2-a e^2\right )}{3 c^3 d^3}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2} \]

[Out]

7/3*e*(-a*e^2+c*d^2)*(e*x+d)^(3/2)/c^3/d^3+7/5*e*(e*x+d)^(5/2)/c^2/d^2-(e*x+d)^(7/2)/c/d/(c*d*x+a*e)-7*e*(-a*e
^2+c*d^2)^(5/2)*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/c^(9/2)/d^(9/2)+7*e*(-a*e^2+c*d^2)
^2*(e*x+d)^(1/2)/c^4/d^4

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Rubi [A]  time = 0.13, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {626, 47, 50, 63, 208} \[ \frac {7 e (d+e x)^{3/2} \left (c d^2-a e^2\right )}{3 c^3 d^3}+\frac {7 e \sqrt {d+e x} \left (c d^2-a e^2\right )^2}{c^4 d^4}-\frac {7 e \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*e*(c*d^2 - a*e^2)^2*Sqrt[d + e*x])/(c^4*d^4) + (7*e*(c*d^2 - a*e^2)*(d + e*x)^(3/2))/(3*c^3*d^3) + (7*e*(d
+ e*x)^(5/2))/(5*c^2*d^2) - (d + e*x)^(7/2)/(c*d*(a*e + c*d*x)) - (7*e*(c*d^2 - a*e^2)^(5/2)*ArcTanh[(Sqrt[c]*
Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c^(9/2)*d^(9/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 626

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac {(d+e x)^{7/2}}{(a e+c d x)^2} \, dx\\ &=-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{a e+c d x} \, dx}{2 c d}\\ &=\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {\left (7 e \left (c d^2-a e^2\right )\right ) \int \frac {(d+e x)^{3/2}}{a e+c d x} \, dx}{2 c^2 d^2}\\ &=\frac {7 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^3 d^3}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {\left (7 e \left (c d^2-a e^2\right )^2\right ) \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{2 c^3 d^3}\\ &=\frac {7 e \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^4 d^4}+\frac {7 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^3 d^3}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {\left (7 e \left (c d^2-a e^2\right )^3\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{2 c^4 d^4}\\ &=\frac {7 e \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^4 d^4}+\frac {7 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^3 d^3}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}+\frac {\left (7 \left (c d^2-a e^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c d^2}{e}+a e+\frac {c d x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{c^4 d^4}\\ &=\frac {7 e \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^4 d^4}+\frac {7 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^3 d^3}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}-\frac {7 e \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 59, normalized size = 0.33 \[ \frac {2 e (d+e x)^{9/2} \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{9 \left (a e^2-c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(d + e*x)^(9/2)*Hypergeometric2F1[2, 9/2, 11/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(9*(-(c*d^2) + a*
e^2)^2)

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fricas [A]  time = 1.07, size = 586, normalized size = 3.29 \[ \left [\frac {105 \, {\left (a c^{2} d^{4} e^{2} - 2 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (6 \, c^{3} d^{3} e^{3} x^{3} - 15 \, c^{3} d^{6} + 161 \, a c^{2} d^{4} e^{2} - 245 \, a^{2} c d^{2} e^{4} + 105 \, a^{3} e^{6} + 2 \, {\left (16 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (58 \, c^{3} d^{5} e - 84 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (c^{5} d^{5} x + a c^{4} d^{4} e\right )}}, -\frac {105 \, {\left (a c^{2} d^{4} e^{2} - 2 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (6 \, c^{3} d^{3} e^{3} x^{3} - 15 \, c^{3} d^{6} + 161 \, a c^{2} d^{4} e^{2} - 245 \, a^{2} c d^{2} e^{4} + 105 \, a^{3} e^{6} + 2 \, {\left (16 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (58 \, c^{3} d^{5} e - 84 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (c^{5} d^{5} x + a c^{4} d^{4} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

[1/30*(105*(a*c^2*d^4*e^2 - 2*a^2*c*d^2*e^4 + a^3*e^6 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x)*sqrt((c
*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x
+ a*e)) + 2*(6*c^3*d^3*e^3*x^3 - 15*c^3*d^6 + 161*a*c^2*d^4*e^2 - 245*a^2*c*d^2*e^4 + 105*a^3*e^6 + 2*(16*c^3*
d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 + 2*(58*c^3*d^5*e - 84*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^5*d
^5*x + a*c^4*d^4*e), -1/15*(105*(a*c^2*d^4*e^2 - 2*a^2*c*d^2*e^4 + a^3*e^6 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^
2*c*d*e^5)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c*d*sqrt(-(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e
^2)) - (6*c^3*d^3*e^3*x^3 - 15*c^3*d^6 + 161*a*c^2*d^4*e^2 - 245*a^2*c*d^2*e^4 + 105*a^3*e^6 + 2*(16*c^3*d^4*e
^2 - 7*a*c^2*d^2*e^4)*x^2 + 2*(58*c^3*d^5*e - 84*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^5*d^5*x
+ a*c^4*d^4*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.07, size = 457, normalized size = 2.57 \[ -\frac {7 a^{3} e^{7} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{4} d^{4}}+\frac {21 a^{2} e^{5} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{3} d^{2}}-\frac {21 a \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{2}}+\frac {7 d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c}+\frac {\sqrt {e x +d}\, a^{3} e^{7}}{\left (c d e x +a \,e^{2}\right ) c^{4} d^{4}}-\frac {3 \sqrt {e x +d}\, a^{2} e^{5}}{\left (c d e x +a \,e^{2}\right ) c^{3} d^{2}}+\frac {3 \sqrt {e x +d}\, a \,e^{3}}{\left (c d e x +a \,e^{2}\right ) c^{2}}-\frac {\sqrt {e x +d}\, d^{2} e}{\left (c d e x +a \,e^{2}\right ) c}+\frac {6 \sqrt {e x +d}\, a^{2} e^{5}}{c^{4} d^{4}}-\frac {12 \sqrt {e x +d}\, a \,e^{3}}{c^{3} d^{2}}+\frac {6 \sqrt {e x +d}\, e}{c^{2}}-\frac {4 \left (e x +d \right )^{\frac {3}{2}} a \,e^{3}}{3 c^{3} d^{3}}+\frac {4 \left (e x +d \right )^{\frac {3}{2}} e}{3 c^{2} d}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} e}{5 c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*e*(e*x+d)^(5/2)/c^2/d^2-4/3/c^3/d^3*(e*x+d)^(3/2)*a*e^3+4/3*e/c^2/d*(e*x+d)^(3/2)+6/c^4/d^4*a^2*e^5*(e*x+d
)^(1/2)-12/c^3/d^2*a*e^3*(e*x+d)^(1/2)+6*e/c^2*(e*x+d)^(1/2)+1/c^4/d^4*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)*a^3*e^7-3
/c^3/d^2*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)*a^2*e^5+3/c^2*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)*a*e^3-e/c*d^2*(e*x+d)^(1/2)
/(c*d*e*x+a*e^2)-7/c^4/d^4/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)*a^3*e
^7+21/c^3/d^2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)*a^2*e^5-21/c^2/((a
*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)*a*e^3+7*e/c*d^2/((a*e^2-c*d^2)*c*d)
^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 positive or negative?

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mupad [B]  time = 0.68, size = 290, normalized size = 1.63 \[ \frac {\sqrt {d+e\,x}\,\left (a^3\,e^7-3\,a^2\,c\,d^2\,e^5+3\,a\,c^2\,d^4\,e^3-c^3\,d^6\,e\right )}{c^5\,d^5\,\left (d+e\,x\right )-c^5\,d^6+a\,c^4\,d^4\,e^2}-\left (\frac {2\,e\,{\left (a\,e^2-c\,d^2\right )}^2}{c^4\,d^4}-\frac {2\,e\,{\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )}^2}{c^6\,d^6}\right )\,\sqrt {d+e\,x}+\frac {2\,e\,{\left (d+e\,x\right )}^{5/2}}{5\,c^2\,d^2}+\frac {2\,e\,\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,c^4\,d^4}-\frac {7\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e\,{\left (a\,e^2-c\,d^2\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^7-3\,a^2\,c\,d^2\,e^5+3\,a\,c^2\,d^4\,e^3-c^3\,d^6\,e}\right )\,{\left (a\,e^2-c\,d^2\right )}^{5/2}}{c^{9/2}\,d^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d + e*x)^(1/2)*(a^3*e^7 - c^3*d^6*e + 3*a*c^2*d^4*e^3 - 3*a^2*c*d^2*e^5))/(c^5*d^5*(d + e*x) - c^5*d^6 + a*c
^4*d^4*e^2) - ((2*e*(a*e^2 - c*d^2)^2)/(c^4*d^4) - (2*e*(2*c^2*d^3 - 2*a*c*d*e^2)^2)/(c^6*d^6))*(d + e*x)^(1/2
) + (2*e*(d + e*x)^(5/2))/(5*c^2*d^2) + (2*e*(2*c^2*d^3 - 2*a*c*d*e^2)*(d + e*x)^(3/2))/(3*c^4*d^4) - (7*e*ata
n((c^(1/2)*d^(1/2)*e*(a*e^2 - c*d^2)^(5/2)*(d + e*x)^(1/2))/(a^3*e^7 - c^3*d^6*e + 3*a*c^2*d^4*e^3 - 3*a^2*c*d
^2*e^5))*(a*e^2 - c*d^2)^(5/2))/(c^(9/2)*d^(9/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(11/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)

[Out]

Timed out

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