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

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

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Rubi [A]  time = 0.20, 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 = {413, 528, 388, 217, 206} \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} \left (c+d x^2\right ) (4 b c-5 a d)}{4 a b^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 {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]

-(d*(2*b*c - 5*a*d)*(4*b*c - 3*a*d)*x*Sqrt[a + b*x^2])/(8*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 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 217

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 388

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 413

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 528

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 ) \operatorname {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*}

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Mathematica [A]  time = 5.10, size = 122, normalized size = 0.72 \begin {gather*} \frac {3 d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{8 b^{7/2}}+\frac {x \sqrt {a+b x^2} \left (d^2 (12 b c-7 a d)+\frac {8 (b c-a d)^3}{a \left (a+b x^2\right )}+2 b d^3 x^2\right )}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 156, normalized size = 0.92 \begin {gather*} \frac {-15 a^3 d^3 x+36 a^2 b c d^2 x-5 a^2 b d^3 x^3-24 a b^2 c^2 d x+12 a b^2 c d^2 x^3+2 a b^2 d^3 x^5+8 b^3 c^3 x}{8 a b^3 \sqrt {a+b x^2}}-\frac {3 \left (5 a^2 d^3-12 a b c d^2+8 b^2 c^2 d\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{8 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^3*c^3*x - 24*a*b^2*c^2*d*x + 36*a^2*b*c*d^2*x - 15*a^3*d^3*x + 12*a*b^2*c*d^2*x^3 - 5*a^2*b*d^3*x^3 + 2*a
*b^2*d^3*x^5)/(8*a*b^3*Sqrt[a + b*x^2]) - (3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*Log[-(Sqrt[b]*x) + Sqrt[
a + b*x^2]])/(8*b^(7/2))

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fricas [A]  time = 0.89, 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)]

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giac [A]  time = 0.64, 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)

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maple [A]  time = 0.01, size = 219, normalized size = 1.30 \begin {gather*} \frac {d^{3} x^{5}}{4 \sqrt {b \,x^{2}+a}\, b}-\frac {5 a \,d^{3} x^{3}}{8 \sqrt {b \,x^{2}+a}\, b^{2}}+\frac {3 c \,d^{2} x^{3}}{2 \sqrt {b \,x^{2}+a}\, b}-\frac {15 a^{2} d^{3} x}{8 \sqrt {b \,x^{2}+a}\, b^{3}}+\frac {9 a c \,d^{2} x}{2 \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 {15 a^{2} d^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {7}{2}}}-\frac {9 a c \,d^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {5}{2}}}+\frac {3 c^{2} d \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.40, 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)

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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 )}^{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)

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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 {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)

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