3.21.1 \(\int \frac {(d+e x)^{7/2}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [2001]

Optimal. Leaf size=147 \[ \frac {2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2}+\frac {2 (d+e x)^{5/2}}{5 c d}-\frac {2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{7/2} d^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 52, 65, 214} \begin {gather*} -\frac {2 \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^{7/2} d^{7/2}}+\frac {2 \sqrt {d+e x} \left (c d^2-a e^2\right )^2}{c^3 d^3}+\frac {2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{3 c^2 d^2}+\frac {2 (d+e x)^{5/2}}{5 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

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 65

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 214

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

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^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)^{7/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\int \frac {(d+e x)^{5/2}}{a e+c d x} \, dx\\ &=\frac {2 (d+e x)^{5/2}}{5 c d}+\frac {\left (c d^2-a e^2\right ) \int \frac {(d+e x)^{3/2}}{a e+c d x} \, dx}{c d}\\ &=\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2}+\frac {2 (d+e x)^{5/2}}{5 c d}+\frac {\left (c d^2-a e^2\right )^2 \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{c^2 d^2}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2}+\frac {2 (d+e x)^{5/2}}{5 c d}+\frac {\left (c d^2-a e^2\right )^3 \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{c^3 d^3}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2}+\frac {2 (d+e x)^{5/2}}{5 c d}+\frac {\left (2 \left (c d^2-a e^2\right )^3\right ) \text {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^3 d^3 e}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2}+\frac {2 (d+e x)^{5/2}}{5 c d}-\frac {2 \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^{7/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 135, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a^2 e^4-5 a c d e^2 (7 d+e x)+c^2 d^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )}{15 c^3 d^3}-\frac {2 \left (-c d^2+a e^2\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{7/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(15*a^2*e^4 - 5*a*c*d*e^2*(7*d + e*x) + c^2*d^2*(23*d^2 + 11*d*e*x + 3*e^2*x^2)))/(15*c^3*d^3
) - (2*(-(c*d^2) + a*e^2)^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(c^(7/2)*d^(7/
2))

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Maple [A]
time = 0.75, size = 194, normalized size = 1.32

method	result	size
derivativedivides	\(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a^{2} e^{4} \sqrt {e x +d}-4 a c \,d^{2} e^{2} \sqrt {e x +d}+2 c^{2} d^{4} \sqrt {e x +d}}{c^{3} d^{3}}+\frac {2 \left (-e^{6} a^{3}+3 e^{4} d^{2} a^{2} c -3 d^{4} e^{2} c^{2} a +d^{6} c^{3}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{3} d^{3} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(194\)
default	\(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a^{2} e^{4} \sqrt {e x +d}-4 a c \,d^{2} e^{2} \sqrt {e x +d}+2 c^{2} d^{4} \sqrt {e x +d}}{c^{3} d^{3}}+\frac {2 \left (-e^{6} a^{3}+3 e^{4} d^{2} a^{2} c -3 d^{4} e^{2} c^{2} a +d^{6} c^{3}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{3} d^{3} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(194\)
risch	\(\frac {2 \left (3 e^{2} x^{2} c^{2} d^{2}-5 a c d \,e^{3} x +11 c^{2} d^{3} e x +15 a^{2} e^{4}-35 a c \,d^{2} e^{2}+23 c^{2} d^{4}\right ) \sqrt {e x +d}}{15 c^{3} d^{3}}-\frac {2 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right ) e^{6} a^{3}}{c^{3} d^{3} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}+\frac {6 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right ) e^{4} a^{2}}{c^{2} d \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}-\frac {6 d \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right ) e^{2} a}{c \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}+\frac {2 d^{3} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(300\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 2.47, size = 354, normalized size = 2.41 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d x e + 2 \, c d^{2} - 2 \, \sqrt {x e + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}} - a e^{2}}{c d x + a e}\right ) + 2 \, {\left (11 \, c^{2} d^{3} x e + 23 \, c^{2} d^{4} - 5 \, a c d x e^{3} + 15 \, a^{2} e^{4} + {\left (3 \, c^{2} d^{2} x^{2} - 35 \, a c d^{2}\right )} e^{2}\right )} \sqrt {x e + d}}{15 \, c^{3} d^{3}}, -\frac {2 \, {\left (15 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {x e + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (11 \, c^{2} d^{3} x e + 23 \, c^{2} d^{4} - 5 \, a c d x e^{3} + 15 \, a^{2} e^{4} + {\left (3 \, c^{2} d^{2} x^{2} - 35 \, a c d^{2}\right )} e^{2}\right )} \sqrt {x e + d}\right )}}{15 \, c^{3} d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 117.43, size = 153, normalized size = 1.04 \begin {gather*} \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 c d} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 2 a e^{2} + 2 c d^{2}\right )}{3 c^{2} d^{2}} + \frac {\sqrt {d + e x} \left (2 a^{2} e^{4} - 4 a c d^{2} e^{2} + 2 c^{2} d^{4}\right )}{c^{3} d^{3}} - \frac {2 \left (a e^{2} - c d^{2}\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{c^{4} d^{4} \sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d + e*x)**(5/2)/(5*c*d) + (d + e*x)**(3/2)*(-2*a*e**2 + 2*c*d**2)/(3*c**2*d**2) + sqrt(d + e*x)*(2*a**2*e**
4 - 4*a*c*d**2*e**2 + 2*c**2*d**4)/(c**3*d**3) - 2*(a*e**2 - c*d**2)**3*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d*
*2)/(c*d)))/(c**4*d**4*sqrt((a*e**2 - c*d**2)/(c*d)))

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Giac [A]
time = 3.12, size = 208, normalized size = 1.41 \begin {gather*} \frac {2 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c^{3} d^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} c^{4} d^{4} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} c^{4} d^{5} + 15 \, \sqrt {x e + d} c^{4} d^{6} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} a c^{3} d^{3} e^{2} - 30 \, \sqrt {x e + d} a c^{3} d^{4} e^{2} + 15 \, \sqrt {x e + d} a^{2} c^{2} d^{2} e^{4}\right )}}{15 \, c^{5} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*arctan(sqrt(x*e + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))
/(sqrt(-c^2*d^3 + a*c*d*e^2)*c^3*d^3) + 2/15*(3*(x*e + d)^(5/2)*c^4*d^4 + 5*(x*e + d)^(3/2)*c^4*d^5 + 15*sqrt(
x*e + d)*c^4*d^6 - 5*(x*e + d)^(3/2)*a*c^3*d^3*e^2 - 30*sqrt(x*e + d)*a*c^3*d^4*e^2 + 15*sqrt(x*e + d)*a^2*c^2
*d^2*e^4)/(c^5*d^5)

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Mupad [B]
time = 0.65, size = 165, normalized size = 1.12 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{5/2}}{5\,c\,d}+\frac {2\,{\left (a\,e^2-c\,d^2\right )}^2\,\sqrt {d+e\,x}}{c^3\,d^3}-\frac {2\,\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,c^2\,d^2}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,{\left (a\,e^2-c\,d^2\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^6-3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2-c^3\,d^6}\right )\,{\left (a\,e^2-c\,d^2\right )}^{5/2}}{c^{7/2}\,d^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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