3.21.42 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{7/2}} \, dx\) [2042]

Optimal. Leaf size=175 \[ \frac {3 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^2 \sqrt {d+e x}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\frac {3 c d \sqrt {c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{e^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {676, 678, 674, 211} \begin {gather*} -\frac {3 c d \sqrt {c d^2-a e^2} \text {ArcTan}\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^{5/2}}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}+\frac {3 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

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

Rule 674

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 676

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 678

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

________________________________________________________________________________________

Mathematica [A]
time = 0.27, size = 142, normalized size = 0.81 \begin {gather*} \frac {\sqrt {(a e+c d x) (d+e x)} \left (\sqrt {e} \sqrt {a e+c d x} \left (-a e^2+c d (3 d+2 e x)\right )-3 c d \sqrt {c d^2-a e^2} (d+e x) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )\right )}{e^{5/2} \sqrt {a e+c d x} (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.75, size = 304, normalized size = 1.74

method	result	size
default	\(\frac {\left (-3 \arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) a c d \,e^{3} x +3 \arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c^{2} d^{3} e x -3 \arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) a c \,d^{2} e^{2}+3 \arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c^{2} d^{4}+2 c d e x \sqrt {c d x +a e}\, \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}-\sqrt {c d x +a e}\, \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\, a \,e^{2}+3 \sqrt {c d x +a e}\, \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\, c \,d^{2}\right ) \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}}{\left (e x +d \right )^{\frac {3}{2}} \sqrt {c d x +a e}\, e^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\)	\(304\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)^(3/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)/(x*e + d)^(7/2), x)

________________________________________________________________________________________

Fricas [A]
time = 2.70, size = 396, normalized size = 2.26 \begin {gather*} \left [\frac {3 \, {\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {-{\left (c d^{2} - a e^{2}\right )} e^{\left (-1\right )}} \log \left (\frac {c d^{3} - 2 \, a x e^{3} + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d} \sqrt {-{\left (c d^{2} - a e^{2}\right )} e^{\left (-1\right )}} e - {\left (c d x^{2} + 2 \, a d\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + 3 \, c d^{2} - a e^{2}\right )} \sqrt {x e + d}}{2 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}}, \frac {3 \, {\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {c d^{2} - a e^{2}} \sqrt {x e + d} e^{\left (-\frac {1}{2}\right )}}{\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}\right ) e^{\left (-\frac {1}{2}\right )} + \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + 3 \, c d^{2} - a e^{2}\right )} \sqrt {x e + d}}{x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}}\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)^(3/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \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)**(3/2)/(e*x+d)**(7/2),x)

[Out]

Integral(((d + e*x)*(a*e + c*d*x))**(3/2)/(d + e*x)**(7/2), x)

________________________________________________________________________________________

Giac [A]
time = 1.24, size = 215, normalized size = 1.23 \begin {gather*} \frac {{\left (2 \, \sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} c^{2} d^{2} - \frac {3 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} \arctan \left (\frac {\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right )}{\sqrt {c d^{2} e - a e^{3}}} + \frac {{\left (\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} c^{3} d^{4} e - \sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} a c^{2} d^{2} e^{3}\right )} e^{\left (-1\right )}}{{\left (x e + d\right )} c d}\right )} e^{\left (-3\right )}}{c d} \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)^(3/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{{\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)^(3/2)/(d + e*x)^(7/2),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2)/(d + e*x)^(7/2), x)

________________________________________________________________________________________