3.1.82 \(\int \frac {(c+d x^2)^3}{(a+b x^2)^{3/2}} \, dx\) [82]

Optimal. Leaf size=169 \[ -\frac {d (2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{8 a b^3}-\frac {d (4 b c-5 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \]

[Out]

3/8*d*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(7/2)+(-a*d+b*c)*x*(d*x^2+c)^2/a/b
/(b*x^2+a)^(1/2)-1/8*d*(-5*a*d+2*b*c)*(-3*a*d+4*b*c)*x*(b*x^2+a)^(1/2)/a/b^3-1/4*d*(-5*a*d+4*b*c)*x*(d*x^2+c)*
(b*x^2+a)^(1/2)/a/b^2

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 169, 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, 542, 396, 223, 212} \begin {gather*} \frac {3 d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}}-\frac {d x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{8 a b^3}-\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right ) (4 b c-5 a d)}{4 a b^2}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(d*(2*b*c - 5*a*d)*(4*b*c - 3*a*d)*x*Sqrt[a + b*x^2])/(a*b^3) - (d*(4*b*c - 5*a*d)*x*Sqrt[a + b*x^2]*(c +
d*x^2))/(4*a*b^2) + ((b*c - a*d)*x*(c + d*x^2)^2)/(a*b*Sqrt[a + b*x^2]) + (3*d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^
2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*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 542

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {\int \frac {\left (c+d x^2\right ) \left (a c d-d (4 b c-5 a d) x^2\right )}{\sqrt {a+b x^2}} \, dx}{a b}\\ &=-\frac {d (4 b c-5 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {\int \frac {a c d (8 b c-5 a d)-d (2 b c-5 a d) (4 b c-3 a d) x^2}{\sqrt {a+b x^2}} \, dx}{4 a b^2}\\ &=-\frac {d (2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{8 a b^3}-\frac {d (4 b c-5 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {\left (3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b^3}\\ &=-\frac {d (2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{8 a b^3}-\frac {d (4 b c-5 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {\left (3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b^3}\\ &=-\frac {d (2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{8 a b^3}-\frac {d (4 b c-5 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.27, size = 139, normalized size = 0.82 \begin {gather*} \frac {x \left (8 b^3 c^3-15 a^3 d^3+a^2 b d^2 \left (36 c-5 d x^2\right )+2 a b^2 d \left (-12 c^2+6 c d x^2+d^2 x^4\right )\right )}{8 a b^3 \sqrt {a+b x^2}}-\frac {3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(8*b^3*c^3 - 15*a^3*d^3 + a^2*b*d^2*(36*c - 5*d*x^2) + 2*a*b^2*d*(-12*c^2 + 6*c*d*x^2 + d^2*x^4)))/(8*a*b^3
*Sqrt[a + b*x^2]) - (3*d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(7/2))

________________________________________________________________________________________

Maple [A]
time = 0.08, size = 215, normalized size = 1.27

method	result	size
risch	\(-\frac {d^{2} x \left (-2 b d \,x^{2}+7 a d -12 b c \right ) \sqrt {b \,x^{2}+a}}{8 b^{3}}-\frac {x \,a^{2} d^{3}}{b^{3} \sqrt {b \,x^{2}+a}}+\frac {3 x a c \,d^{2}}{b^{2} \sqrt {b \,x^{2}+a}}-\frac {3 x \,c^{2} d}{b \sqrt {b \,x^{2}+a}}+\frac {15 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{2} d^{3}}{8 b^{\frac {7}{2}}}-\frac {9 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a c \,d^{2}}{2 b^{\frac {5}{2}}}+\frac {3 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c^{2} d}{b^{\frac {3}{2}}}+\frac {c^{3} x}{a \sqrt {b \,x^{2}+a}}\)	\(191\)
default	\(d^{3} \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )+3 c \,d^{2} \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+3 c^{2} d \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+\frac {c^{3} x}{a \sqrt {b \,x^{2}+a}}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^3*(1/4*x^5/b/(b*x^2+a)^(1/2)-5/4*a/b*(1/2*x^3/b/(b*x^2+a)^(1/2)-3/2*a/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*(1/2*x^3/b/(b*x^2+a)^(1/2)-3/2*a/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*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))+c^3*x/a/(b*x^
2+a)^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 197, normalized size = 1.17 \begin {gather*} \frac {d^{3} x^{5}}{4 \, \sqrt {b x^{2} + a} b} + \frac {3 \, c d^{2} x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {5 \, a d^{3} x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {c^{3} x}{\sqrt {b x^{2} + a} a} - \frac {3 \, c^{2} d x}{\sqrt {b x^{2} + a} b} + \frac {9 \, a c d^{2} x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {15 \, a^{2} d^{3} x}{8 \, \sqrt {b x^{2} + a} b^{3}} + \frac {3 \, c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {9 \, a c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {15 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/4*d^3*x^5/(sqrt(b*x^2 + a)*b) + 3/2*c*d^2*x^3/(sqrt(b*x^2 + a)*b) - 5/8*a*d^3*x^3/(sqrt(b*x^2 + a)*b^2) + c^
3*x/(sqrt(b*x^2 + a)*a) - 3*c^2*d*x/(sqrt(b*x^2 + a)*b) + 9/2*a*c*d^2*x/(sqrt(b*x^2 + a)*b^2) - 15/8*a^2*d^3*x
/(sqrt(b*x^2 + a)*b^3) + 3*c^2*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) - 9/2*a*c*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) +
15/8*a^2*d^3*arcsinh(b*x/sqrt(a*b))/b^(7/2)

________________________________________________________________________________________

Fricas [A]
time = 0.54, size = 416, normalized size = 2.46 \begin {gather*} \left [\frac {3 \, {\left (8 \, a^{2} b^{2} c^{2} d - 12 \, a^{3} b c d^{2} + 5 \, a^{4} d^{3} + {\left (8 \, a b^{3} c^{2} d - 12 \, a^{2} b^{2} c d^{2} + 5 \, a^{3} 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 (2 \, a b^{3} d^{3} x^{5} + {\left (12 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x^{3} + {\left (8 \, b^{4} c^{3} - 24 \, a b^{3} c^{2} d + 36 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}, -\frac {3 \, {\left (8 \, a^{2} b^{2} c^{2} d - 12 \, a^{3} b c d^{2} + 5 \, a^{4} d^{3} + {\left (8 \, a b^{3} c^{2} d - 12 \, a^{2} b^{2} c d^{2} + 5 \, a^{3} b d^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, a b^{3} d^{3} x^{5} + {\left (12 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x^{3} + {\left (8 \, b^{4} c^{3} - 24 \, a b^{3} c^{2} d + 36 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, {\left (a b^{5} x^{2} + a^{2} 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)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*(8*a^2*b^2*c^2*d - 12*a^3*b*c*d^2 + 5*a^4*d^3 + (8*a*b^3*c^2*d - 12*a^2*b^2*c*d^2 + 5*a^3*b*d^3)*x^2)
*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*a*b^3*d^3*x^5 + (12*a*b^3*c*d^2 - 5*a^2*b^2*d^
3)*x^3 + (8*b^4*c^3 - 24*a*b^3*c^2*d + 36*a^2*b^2*c*d^2 - 15*a^3*b*d^3)*x)*sqrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b
^4), -1/8*(3*(8*a^2*b^2*c^2*d - 12*a^3*b*c*d^2 + 5*a^4*d^3 + (8*a*b^3*c^2*d - 12*a^2*b^2*c*d^2 + 5*a^3*b*d^3)*
x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*a*b^3*d^3*x^5 + (12*a*b^3*c*d^2 - 5*a^2*b^2*d^3)*x^3 + (
8*b^4*c^3 - 24*a*b^3*c^2*d + 36*a^2*b^2*c*d^2 - 15*a^3*b*d^3)*x)*sqrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b^4)]

________________________________________________________________________________________

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 {3}{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)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.57, size = 157, normalized size = 0.93 \begin {gather*} \frac {{\left ({\left (\frac {2 \, d^{3} x^{2}}{b} + \frac {12 \, a b^{4} c d^{2} - 5 \, a^{2} b^{3} d^{3}}{a b^{5}}\right )} x^{2} + \frac {8 \, b^{5} c^{3} - 24 \, a b^{4} c^{2} d + 36 \, a^{2} b^{3} c d^{2} - 15 \, a^{3} b^{2} d^{3}}{a b^{5}}\right )} x}{8 \, \sqrt {b x^{2} + a}} - \frac {3 \, {\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, 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)^(3/2),x, algorithm="giac")

[Out]

1/8*((2*d^3*x^2/b + (12*a*b^4*c*d^2 - 5*a^2*b^3*d^3)/(a*b^5))*x^2 + (8*b^5*c^3 - 24*a*b^4*c^2*d + 36*a^2*b^3*c
*d^2 - 15*a^3*b^2*d^3)/(a*b^5))*x/sqrt(b*x^2 + a) - 3/8*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*log(abs(-sqrt
(b)*x + sqrt(b*x^2 + a)))/b^(7/2)

________________________________________________________________________________________

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 )}^{3/2}} \,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)^(3/2),x)

[Out]

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

________________________________________________________________________________________