3.1.90 \(\int \frac {(c+d x^2)^3}{(a+b x^2)^{5/2}} \, dx\) [90]

Optimal. Leaf size=172 \[ -\frac {d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt {a+b x^2}}{6 a^2 b^3}+\frac {(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {d^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \]

[Out]

1/3*(-a*d+b*c)*x*(d*x^2+c)^2/a/b/(b*x^2+a)^(3/2)+1/2*d^2*(-5*a*d+6*b*c)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(
7/2)+1/3*(-a*d+b*c)*(5*a*d+2*b*c)*x*(d*x^2+c)/a^2/b^2/(b*x^2+a)^(1/2)-1/6*d*(-15*a^2*d^2+8*a*b*c*d+4*b^2*c^2)*
x*(b*x^2+a)^(1/2)/a^2/b^3

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Rubi [A]
time = 0.10, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {424, 540, 396, 223, 212} \begin {gather*} \frac {x \left (c+d x^2\right ) (b c-a d) (5 a d+2 b c)}{3 a^2 b^2 \sqrt {a+b x^2}}-\frac {d x \sqrt {a+b x^2} \left (-15 a^2 d^2+8 a b c d+4 b^2 c^2\right )}{6 a^2 b^3}+\frac {d^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^3/(a + b*x^2)^(5/2),x]

[Out]

-1/6*(d*(4*b^2*c^2 + 8*a*b*c*d - 15*a^2*d^2)*x*Sqrt[a + b*x^2])/(a^2*b^3) + ((b*c - a*d)*(2*b*c + 5*a*d)*x*(c
+ d*x^2))/(3*a^2*b^2*Sqrt[a + b*x^2]) + ((b*c - a*d)*x*(c + d*x^2)^2)/(3*a*b*(a + b*x^2)^(3/2)) + (d^2*(6*b*c
- 5*a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(7/2))

Rule 212

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

Rule 223

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

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 540

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*n*(p + 1))), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x
^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))
*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rubi steps

\begin {align*} \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac {(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {\int \frac {\left (c+d x^2\right ) \left (c (2 b c+a d)-d (2 b c-5 a d) x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a b}\\ &=\frac {(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {a c d (2 b c-5 a d)+d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x^2}{\sqrt {a+b x^2}} \, dx}{3 a^2 b^2}\\ &=-\frac {d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt {a+b x^2}}{6 a^2 b^3}+\frac {(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {\left (d^2 (6 b c-5 a d)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{2 b^3}\\ &=-\frac {d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt {a+b x^2}}{6 a^2 b^3}+\frac {(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {\left (d^2 (6 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{2 b^3}\\ &=-\frac {d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt {a+b x^2}}{6 a^2 b^3}+\frac {(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {d^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 143, normalized size = 0.83 \begin {gather*} \frac {x \left (15 a^4 d^3+4 b^4 c^3 x^2+3 a^2 b^2 d^2 x^2 \left (-8 c+d x^2\right )+6 a b^3 c^2 \left (c+d x^2\right )+2 a^3 b d^2 \left (-9 c+10 d x^2\right )\right )}{6 a^2 b^3 \left (a+b x^2\right )^{3/2}}+\frac {d^2 (-6 b c+5 a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^3/(a + b*x^2)^(5/2),x]

[Out]

(x*(15*a^4*d^3 + 4*b^4*c^3*x^2 + 3*a^2*b^2*d^2*x^2*(-8*c + d*x^2) + 6*a*b^3*c^2*(c + d*x^2) + 2*a^3*b*d^2*(-9*
c + 10*d*x^2)))/(6*a^2*b^3*(a + b*x^2)^(3/2)) + (d^2*(-6*b*c + 5*a*d)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*
b^(7/2))

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Maple [A]
time = 0.10, size = 246, normalized size = 1.43

method	result	size
default	\(d^{3} \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )+3 c \,d^{2} \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+3 c^{2} d \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )+c^{3} \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(246\)
risch	\(\frac {d^{3} x \sqrt {b \,x^{2}+a}}{2 b^{3}}+\frac {3 d^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c}{b^{\frac {5}{2}}}-\frac {5 d^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a}{2 b^{\frac {7}{2}}}+\frac {a^{2} \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, d^{3}}{12 b^{4} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, c^{3}}{12 b a \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {2 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, c \,d^{2}}{b^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, c^{2} d}{2 b^{2} a \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {a^{2} \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, d^{3}}{12 b^{4} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, c^{3}}{12 b a \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {2 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, c \,d^{2}}{b^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, c^{2} d}{2 b^{2} a \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {a \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, c \,d^{2}}{4 b^{3} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, c^{2} d}{4 b^{2} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {a \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, c \,d^{2}}{4 b^{3} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, c^{2} d}{4 b^{2} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {7 a \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, d^{3}}{6 b^{4} \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, c^{3}}{3 b \,a^{2} \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {7 a \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, d^{3}}{6 b^{4} \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, c^{3}}{3 b \,a^{2} \left (x -\frac {\sqrt {-a b}}{b}\right )}\)	\(1154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^3/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

d^3*(1/2*x^5/b/(b*x^2+a)^(3/2)-5/2*a/b*(-1/3*x^3/b/(b*x^2+a)^(3/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(x*b^
(1/2)+(b*x^2+a)^(1/2)))))+3*c*d^2*(-1/3*x^3/b/(b*x^2+a)^(3/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(x*b^(1/2)
+(b*x^2+a)^(1/2))))+3*c^2*d*(-1/2*x/b/(b*x^2+a)^(3/2)+1/2*a/b*(1/3*x/a/(b*x^2+a)^(3/2)+2/3*x/a^2/(b*x^2+a)^(1/
2)))+c^3*(1/3*x/a/(b*x^2+a)^(3/2)+2/3*x/a^2/(b*x^2+a)^(1/2))

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Maxima [A]
time = 0.30, size = 254, normalized size = 1.48 \begin {gather*} \frac {d^{3} x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - c d^{2} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {5 \, a d^{3} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {2 \, c^{3} x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {c^{3} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {c^{2} d x}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {c^{2} d x}{\sqrt {b x^{2} + a} a b} - \frac {c d^{2} x}{\sqrt {b x^{2} + a} b^{2}} + \frac {5 \, a d^{3} x}{6 \, \sqrt {b x^{2} + a} b^{3}} + \frac {3 \, c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {5 \, a d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/2*d^3*x^5/((b*x^2 + a)^(3/2)*b) - c*d^2*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) + 5/6*
a*d^3*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2))/b + 2/3*c^3*x/(sqrt(b*x^2 + a)*a^2) + 1/3*
c^3*x/((b*x^2 + a)^(3/2)*a) - c^2*d*x/((b*x^2 + a)^(3/2)*b) + c^2*d*x/(sqrt(b*x^2 + a)*a*b) - c*d^2*x/(sqrt(b*
x^2 + a)*b^2) + 5/6*a*d^3*x/(sqrt(b*x^2 + a)*b^3) + 3*c*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 5/2*a*d^3*arcsinh
(b*x/sqrt(a*b))/b^(7/2)

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Fricas [A]
time = 0.58, size = 486, normalized size = 2.83 \begin {gather*} \left [-\frac {3 \, {\left (6 \, a^{4} b c d^{2} - 5 \, a^{5} d^{3} + {\left (6 \, a^{2} b^{3} c d^{2} - 5 \, a^{3} b^{2} d^{3}\right )} x^{4} + 2 \, {\left (6 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, a^{2} b^{3} d^{3} x^{5} + 2 \, {\left (2 \, b^{5} c^{3} + 3 \, a b^{4} c^{2} d - 12 \, a^{2} b^{3} c d^{2} + 10 \, a^{3} b^{2} d^{3}\right )} x^{3} + 3 \, {\left (2 \, a b^{4} c^{3} - 6 \, a^{3} b^{2} c d^{2} + 5 \, a^{4} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}, -\frac {3 \, {\left (6 \, a^{4} b c d^{2} - 5 \, a^{5} d^{3} + {\left (6 \, a^{2} b^{3} c d^{2} - 5 \, a^{3} b^{2} d^{3}\right )} x^{4} + 2 \, {\left (6 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, a^{2} b^{3} d^{3} x^{5} + 2 \, {\left (2 \, b^{5} c^{3} + 3 \, a b^{4} c^{2} d - 12 \, a^{2} b^{3} c d^{2} + 10 \, a^{3} b^{2} d^{3}\right )} x^{3} + 3 \, {\left (2 \, a b^{4} c^{3} - 6 \, a^{3} b^{2} c d^{2} + 5 \, a^{4} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(6*a^4*b*c*d^2 - 5*a^5*d^3 + (6*a^2*b^3*c*d^2 - 5*a^3*b^2*d^3)*x^4 + 2*(6*a^3*b^2*c*d^2 - 5*a^4*b*d^
3)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(3*a^2*b^3*d^3*x^5 + 2*(2*b^5*c^3 + 3*a*b^
4*c^2*d - 12*a^2*b^3*c*d^2 + 10*a^3*b^2*d^3)*x^3 + 3*(2*a*b^4*c^3 - 6*a^3*b^2*c*d^2 + 5*a^4*b*d^3)*x)*sqrt(b*x
^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4), -1/6*(3*(6*a^4*b*c*d^2 - 5*a^5*d^3 + (6*a^2*b^3*c*d^2 - 5*a^
3*b^2*d^3)*x^4 + 2*(6*a^3*b^2*c*d^2 - 5*a^4*b*d^3)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (3*a^2*b
^3*d^3*x^5 + 2*(2*b^5*c^3 + 3*a*b^4*c^2*d - 12*a^2*b^3*c*d^2 + 10*a^3*b^2*d^3)*x^3 + 3*(2*a*b^4*c^3 - 6*a^3*b^
2*c*d^2 + 5*a^4*b*d^3)*x)*sqrt(b*x^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x^{2}\right )^{3}}{\left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**3/(b*x**2+a)**(5/2),x)

[Out]

Integral((c + d*x**2)**3/(a + b*x**2)**(5/2), x)

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Giac [A]
time = 0.65, size = 158, normalized size = 0.92 \begin {gather*} \frac {{\left ({\left (\frac {3 \, d^{3} x^{2}}{b} + \frac {2 \, {\left (2 \, b^{6} c^{3} + 3 \, a b^{5} c^{2} d - 12 \, a^{2} b^{4} c d^{2} + 10 \, a^{3} b^{3} d^{3}\right )}}{a^{2} b^{5}}\right )} x^{2} + \frac {3 \, {\left (2 \, a b^{5} c^{3} - 6 \, a^{3} b^{3} c d^{2} + 5 \, a^{4} b^{2} d^{3}\right )}}{a^{2} b^{5}}\right )} x}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {{\left (6 \, b c d^{2} - 5 \, a d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/6*((3*d^3*x^2/b + 2*(2*b^6*c^3 + 3*a*b^5*c^2*d - 12*a^2*b^4*c*d^2 + 10*a^3*b^3*d^3)/(a^2*b^5))*x^2 + 3*(2*a*
b^5*c^3 - 6*a^3*b^3*c*d^2 + 5*a^4*b^2*d^3)/(a^2*b^5))*x/(b*x^2 + a)^(3/2) - 1/2*(6*b*c*d^2 - 5*a*d^3)*log(abs(
-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^3}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^3/(a + b*x^2)^(5/2),x)

[Out]

int((c + d*x^2)^3/(a + b*x^2)^(5/2), x)

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