3.1252 \(\int \frac {(b d+2 c d x)^4}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ 16 c^{3/2} d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {8 c d^4 (b+2 c x)}{\sqrt {a+b x+c x^2}}-\frac {2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

-2/3*d^4*(2*c*x+b)^3/(c*x^2+b*x+a)^(3/2)+16*c^(3/2)*d^4*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))-8*c
*d^4*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {686, 621, 206} \[ 16 c^{3/2} d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {8 c d^4 (b+2 c x)}{\sqrt {a+b x+c x^2}}-\frac {2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^4*(b + 2*c*x)^3)/(3*(a + b*x + c*x^2)^(3/2)) - (8*c*d^4*(b + 2*c*x))/Sqrt[a + b*x + c*x^2] + 16*c^(3/2)*
d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

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

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(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[2*c*d - b*e, 0] && NeQ[m + 2
*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}+\left (4 c d^2\right ) \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {8 c d^4 (b+2 c x)}{\sqrt {a+b x+c x^2}}+\left (16 c^2 d^4\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {8 c d^4 (b+2 c x)}{\sqrt {a+b x+c x^2}}+\left (32 c^2 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )\\ &=-\frac {2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {8 c d^4 (b+2 c x)}{\sqrt {a+b x+c x^2}}+16 c^{3/2} d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 142, normalized size = 1.48 \[ d^4 \left (\frac {16 c^{3/2} \sqrt {a+x (b+c x)} \sinh ^{-1}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {4 a-\frac {b^2}{c}}}\right )}{\sqrt {4 a-\frac {b^2}{c}} \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}}-\frac {2 (b+2 c x) \left (4 c \left (3 a+4 c x^2\right )+b^2+16 b c x\right )}{3 (a+x (b+c x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(5/2),x]

[Out]

d^4*((-2*(b + 2*c*x)*(b^2 + 16*b*c*x + 4*c*(3*a + 4*c*x^2)))/(3*(a + x*(b + c*x))^(3/2)) + (16*c^(3/2)*Sqrt[a
+ x*(b + c*x)]*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])])/(Sqrt[4*a - b^2/c]*Sqrt[(c*(a + x*(b + c*x)))
/(-b^2 + 4*a*c)]))

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fricas [B]  time = 1.74, size = 440, normalized size = 4.58 \[ \left [\frac {2 \, {\left (12 \, {\left (c^{3} d^{4} x^{4} + 2 \, b c^{2} d^{4} x^{3} + 2 \, a b c d^{4} x + a^{2} c d^{4} + {\left (b^{2} c + 2 \, a c^{2}\right )} d^{4} x^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - {\left (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c + 4 \, a c^{2}\right )} d^{4} x + {\left (b^{3} + 12 \, a b c\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}}, -\frac {2 \, {\left (24 \, {\left (c^{3} d^{4} x^{4} + 2 \, b c^{2} d^{4} x^{3} + 2 \, a b c d^{4} x + a^{2} c d^{4} + {\left (b^{2} c + 2 \, a c^{2}\right )} d^{4} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c + 4 \, a c^{2}\right )} d^{4} x + {\left (b^{3} + 12 \, a b c\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[2/3*(12*(c^3*d^4*x^4 + 2*b*c^2*d^4*x^3 + 2*a*b*c*d^4*x + a^2*c*d^4 + (b^2*c + 2*a*c^2)*d^4*x^2)*sqrt(c)*log(-
8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - (32*c^3*d^4*x^3 + 48*b*c^2*
d^4*x^2 + 6*(3*b^2*c + 4*a*c^2)*d^4*x + (b^3 + 12*a*b*c)*d^4)*sqrt(c*x^2 + b*x + a))/(c^2*x^4 + 2*b*c*x^3 + 2*
a*b*x + (b^2 + 2*a*c)*x^2 + a^2), -2/3*(24*(c^3*d^4*x^4 + 2*b*c^2*d^4*x^3 + 2*a*b*c*d^4*x + a^2*c*d^4 + (b^2*c
+ 2*a*c^2)*d^4*x^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) +
(32*c^3*d^4*x^3 + 48*b*c^2*d^4*x^2 + 6*(3*b^2*c + 4*a*c^2)*d^4*x + (b^3 + 12*a*b*c)*d^4)*sqrt(c*x^2 + b*x + a
))/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)]

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giac [B]  time = 0.31, size = 315, normalized size = 3.28 \[ -16 \, c^{\frac {3}{2}} d^{4} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right ) - \frac {2 \, {\left (2 \, {\left (8 \, {\left (\frac {2 \, {\left (b^{4} c^{3} d^{4} - 8 \, a b^{2} c^{4} d^{4} + 16 \, a^{2} c^{5} d^{4}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (b^{5} c^{2} d^{4} - 8 \, a b^{3} c^{3} d^{4} + 16 \, a^{2} b c^{4} d^{4}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (3 \, b^{6} c d^{4} - 20 \, a b^{4} c^{2} d^{4} + 16 \, a^{2} b^{2} c^{3} d^{4} + 64 \, a^{3} c^{4} d^{4}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {b^{7} d^{4} + 4 \, a b^{5} c d^{4} - 80 \, a^{2} b^{3} c^{2} d^{4} + 192 \, a^{3} b c^{3} d^{4}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-16*c^(3/2)*d^4*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b)) - 2/3*(2*(8*(2*(b^4*c^3*d^4 - 8*a
*b^2*c^4*d^4 + 16*a^2*c^5*d^4)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(b^5*c^2*d^4 - 8*a*b^3*c^3*d^4 + 16*a^2*b*
c^4*d^4)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 3*(3*b^6*c*d^4 - 20*a*b^4*c^2*d^4 + 16*a^2*b^2*c^3*d^4 + 64*a^3*c
^4*d^4)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + (b^7*d^4 + 4*a*b^5*c*d^4 - 80*a^2*b^3*c^2*d^4 + 192*a^3*b*c^3*d^4)
/(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c*x^2 + b*x + a)^(3/2)

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maple [B]  time = 0.06, size = 531, normalized size = 5.53 \[ -\frac {64 a \,b^{2} c^{3} d^{4} x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {16 b^{4} c^{2} d^{4} x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {32 a \,b^{3} c^{2} d^{4}}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {8 a \,b^{2} c^{2} d^{4} x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {8 b^{5} c \,d^{4}}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {2 b^{4} c \,d^{4} x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {16 c^{3} d^{4} x^{3}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {4 a \,b^{3} c \,d^{4}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b^{5} d^{4}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 b^{2} c^{2} d^{4} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {24 b \,c^{2} d^{4} x^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {8 b^{3} c \,d^{4}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {18 b^{2} c \,d^{4} x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {16 a b c \,d^{4}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b^{3} d^{4}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {16 c^{2} d^{4} x}{\sqrt {c \,x^{2}+b x +a}}+16 c^{\frac {3}{2}} d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )+\frac {8 b c \,d^{4}}{\sqrt {c \,x^{2}+b x +a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(5/2),x)

[Out]

-24*d^4*c^2*b*x^2/(c*x^2+b*x+a)^(3/2)-18*d^4*c*b^2*x/(c*x^2+b*x+a)^(3/2)+8*d^4*c*b^5/(4*a*c-b^2)^2/(c*x^2+b*x+
a)^(1/2)-16*d^4*c*b*a/(c*x^2+b*x+a)^(3/2)+8*d^4*c*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-8*d^4*c^2*b^2*a/(4*a*c-b
^2)/(c*x^2+b*x+a)^(3/2)*x-64*d^4*c^3*b^2*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-16/3*d^4*c^3*x^3/(c*x^2+b*x+a)^
(3/2)+d^4*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-16*d^4*c^2*x/(c*x^2+b*x+a)^(1/2)+8*d^4*c*b/(c*x^2+b*x+a)^(1/2)+1
6*d^4*c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*d^4*b^3/(c*x^2+b*x+a)^(3/2)-4*d^4*c*b^3*a/(4*a*c
-b^2)/(c*x^2+b*x+a)^(3/2)-32*d^4*c^2*b^3*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+16*d^4*c^2*b^2/(4*a*c-b^2)/(c*x^2
+b*x+a)^(1/2)*x+2*d^4*c*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+16*d^4*c^2*b^4/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)
*x

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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((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(5/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,d+2\,c\,d\,x\right )}^4}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(5/2),x)

[Out]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(5/2), x)

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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((2*c*d*x+b*d)**4/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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