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

Optimal. Leaf size=255 \[ -\frac{d x \sqrt{a+b x^2} \left (c+d x^2\right ) \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{12 a^2 b^3}-\frac{d x \sqrt{a+b x^2} \left (-170 a^2 b c d^2+105 a^3 d^3+40 a b^2 c^2 d+16 b^3 c^3\right )}{24 a^2 b^4}+\frac{d^2 \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{9/2}}+\frac{x \left (c+d x^2\right )^2 (b c-a d) (7 a d+2 b c)}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{x \left (c+d x^2\right )^3 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \]

[Out]

-(d*(16*b^3*c^3 + 40*a*b^2*c^2*d - 170*a^2*b*c*d^2 + 105*a^3*d^3)*x*Sqrt[a + b*x^2])/(24*a^2*b^4) - (d*(8*b^2*
c^2 + 24*a*b*c*d - 35*a^2*d^2)*x*Sqrt[a + b*x^2]*(c + d*x^2))/(12*a^2*b^3) + ((b*c - a*d)*(2*b*c + 7*a*d)*x*(c
+ d*x^2)^2)/(3*a^2*b^2*Sqrt[a + b*x^2]) + ((b*c - a*d)*x*(c + d*x^2)^3)/(3*a*b*(a + b*x^2)^(3/2)) + (d^2*(48*
b^2*c^2 - 80*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*b^(9/2))

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Rubi [A]  time = 0.244889, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {413, 526, 528, 388, 217, 206} \[ -\frac{d x \sqrt{a+b x^2} \left (c+d x^2\right ) \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{12 a^2 b^3}-\frac{d x \sqrt{a+b x^2} \left (-170 a^2 b c d^2+105 a^3 d^3+40 a b^2 c^2 d+16 b^3 c^3\right )}{24 a^2 b^4}+\frac{d^2 \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{9/2}}+\frac{x \left (c+d x^2\right )^2 (b c-a d) (7 a d+2 b c)}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{x \left (c+d x^2\right )^3 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(16*b^3*c^3 + 40*a*b^2*c^2*d - 170*a^2*b*c*d^2 + 105*a^3*d^3)*x*Sqrt[a + b*x^2])/(24*a^2*b^4) - (d*(8*b^2*
c^2 + 24*a*b*c*d - 35*a^2*d^2)*x*Sqrt[a + b*x^2]*(c + d*x^2))/(12*a^2*b^3) + ((b*c - a*d)*(2*b*c + 7*a*d)*x*(c
+ d*x^2)^2)/(3*a^2*b^2*Sqrt[a + b*x^2]) + ((b*c - a*d)*x*(c + d*x^2)^3)/(3*a*b*(a + b*x^2)^(3/2)) + (d^2*(48*
b^2*c^2 - 80*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*b^(9/2))

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 526

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]

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{\left (c+d x^2\right )^4}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac{(b c-a d) x \left (c+d x^2\right )^3}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{\int \frac{\left (c+d x^2\right )^2 \left (c (2 b c+a d)-d (4 b c-7 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+7 a d) x \left (c+d x^2\right )^2}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^3}{3 a b \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{\left (c+d x^2\right ) \left (a c d (4 b c-7 a d)+d \left (8 b^2 c^2+24 a b c d-35 a^2 d^2\right ) x^2\right )}{\sqrt{a+b x^2}} \, dx}{3 a^2 b^2}\\ &=-\frac{d \left (8 b^2 c^2+24 a b c d-35 a^2 d^2\right ) x \sqrt{a+b x^2} \left (c+d x^2\right )}{12 a^2 b^3}+\frac{(b c-a d) (2 b c+7 a d) x \left (c+d x^2\right )^2}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^3}{3 a b \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{a c d \left (8 b^2 c^2-52 a b c d+35 a^2 d^2\right )+d \left (16 b^3 c^3+40 a b^2 c^2 d-170 a^2 b c d^2+105 a^3 d^3\right ) x^2}{\sqrt{a+b x^2}} \, dx}{12 a^2 b^3}\\ &=-\frac{d \left (16 b^3 c^3+40 a b^2 c^2 d-170 a^2 b c d^2+105 a^3 d^3\right ) x \sqrt{a+b x^2}}{24 a^2 b^4}-\frac{d \left (8 b^2 c^2+24 a b c d-35 a^2 d^2\right ) x \sqrt{a+b x^2} \left (c+d x^2\right )}{12 a^2 b^3}+\frac{(b c-a d) (2 b c+7 a d) x \left (c+d x^2\right )^2}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^3}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{\left (d^2 \left (48 b^2 c^2-80 a b c d+35 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{8 b^4}\\ &=-\frac{d \left (16 b^3 c^3+40 a b^2 c^2 d-170 a^2 b c d^2+105 a^3 d^3\right ) x \sqrt{a+b x^2}}{24 a^2 b^4}-\frac{d \left (8 b^2 c^2+24 a b c d-35 a^2 d^2\right ) x \sqrt{a+b x^2} \left (c+d x^2\right )}{12 a^2 b^3}+\frac{(b c-a d) (2 b c+7 a d) x \left (c+d x^2\right )^2}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^3}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{\left (d^2 \left (48 b^2 c^2-80 a b c d+35 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^4}\\ &=-\frac{d \left (16 b^3 c^3+40 a b^2 c^2 d-170 a^2 b c d^2+105 a^3 d^3\right ) x \sqrt{a+b x^2}}{24 a^2 b^4}-\frac{d \left (8 b^2 c^2+24 a b c d-35 a^2 d^2\right ) x \sqrt{a+b x^2} \left (c+d x^2\right )}{12 a^2 b^3}+\frac{(b c-a d) (2 b c+7 a d) x \left (c+d x^2\right )^2}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^3}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{d^2 \left (48 b^2 c^2-80 a b c d+35 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 5.16222, size = 157, normalized size = 0.62 \[ \frac{d^2 \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{8 b^{9/2}}+\frac{x \sqrt{a+b x^2} \left (\frac{16 (b c-a d)^3 (5 a d+b c)}{a^2 \left (a+b x^2\right )}+3 d^3 (16 b c-11 a d)+\frac{8 (b c-a d)^4}{a \left (a+b x^2\right )^2}+6 b d^4 x^2\right )}{24 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(3*d^3*(16*b*c - 11*a*d) + 6*b*d^4*x^2 + (8*(b*c - a*d)^4)/(a*(a + b*x^2)^2) + (16*(b*c - a
*d)^3*(b*c + 5*a*d))/(a^2*(a + b*x^2))))/(24*b^4) + (d^2*(48*b^2*c^2 - 80*a*b*c*d + 35*a^2*d^2)*Log[b*x + Sqrt
[b]*Sqrt[a + b*x^2]])/(8*b^(9/2))

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Maple [A]  time = 0.02, size = 351, normalized size = 1.4 \begin{align*}{\frac{{d}^{4}{x}^{7}}{4\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,a{d}^{4}{x}^{5}}{8\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{35\,{a}^{2}{d}^{4}{x}^{3}}{24\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{35\,{a}^{2}{d}^{4}x}{8\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{35\,{a}^{2}{d}^{4}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{9}{2}}}}+2\,{\frac{c{d}^{3}{x}^{5}}{b \left ( b{x}^{2}+a \right ) ^{3/2}}}+{\frac{10\,ac{d}^{3}{x}^{3}}{3\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+10\,{\frac{ac{d}^{3}x}{{b}^{3}\sqrt{b{x}^{2}+a}}}-10\,{\frac{ac{d}^{3}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) }{{b}^{7/2}}}-2\,{\frac{{c}^{2}{d}^{2}{x}^{3}}{b \left ( b{x}^{2}+a \right ) ^{3/2}}}-6\,{\frac{{c}^{2}{d}^{2}x}{{b}^{2}\sqrt{b{x}^{2}+a}}}+6\,{\frac{{c}^{2}{d}^{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) }{{b}^{5/2}}}-{\frac{4\,{c}^{3}dx}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,{c}^{3}dx}{3\,ab}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{{c}^{4}x}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{c}^{4}x}{3\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^4*x^7/b/(b*x^2+a)^(3/2)-7/8*d^4/b^2*a*x^5/(b*x^2+a)^(3/2)-35/24*d^4/b^3*a^2*x^3/(b*x^2+a)^(3/2)-35/8*d^4
/b^4*a^2*x/(b*x^2+a)^(1/2)+35/8*d^4/b^(9/2)*a^2*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+2*c*d^3*x^5/b/(b*x^2+a)^(3/2)+10
/3*c*d^3/b^2*a*x^3/(b*x^2+a)^(3/2)+10*c*d^3/b^3*a*x/(b*x^2+a)^(1/2)-10*c*d^3/b^(7/2)*a*ln(x*b^(1/2)+(b*x^2+a)^
(1/2))-2*c^2*d^2*x^3/b/(b*x^2+a)^(3/2)-6*c^2*d^2/b^2*x/(b*x^2+a)^(1/2)+6*c^2*d^2/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a
)^(1/2))-4/3*c^3*d/b*x/(b*x^2+a)^(3/2)+4/3*c^3*d/b/a*x/(b*x^2+a)^(1/2)+1/3*c^4*x/a/(b*x^2+a)^(3/2)+2/3*c^4/a^2
*x/(b*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.78682, size = 1430, normalized size = 5.61 \begin{align*} \left [\frac{3 \,{\left (48 \, a^{4} b^{2} c^{2} d^{2} - 80 \, a^{5} b c d^{3} + 35 \, a^{6} d^{4} +{\left (48 \, a^{2} b^{4} c^{2} d^{2} - 80 \, a^{3} b^{3} c d^{3} + 35 \, a^{4} b^{2} d^{4}\right )} x^{4} + 2 \,{\left (48 \, a^{3} b^{3} c^{2} d^{2} - 80 \, a^{4} b^{2} c d^{3} + 35 \, a^{5} b d^{4}\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 (6 \, a^{2} b^{4} d^{4} x^{7} + 3 \,{\left (16 \, a^{2} b^{4} c d^{3} - 7 \, a^{3} b^{3} d^{4}\right )} x^{5} + 4 \,{\left (4 \, b^{6} c^{4} + 8 \, a b^{5} c^{3} d - 48 \, a^{2} b^{4} c^{2} d^{2} + 80 \, a^{3} b^{3} c d^{3} - 35 \, a^{4} b^{2} d^{4}\right )} x^{3} + 3 \,{\left (8 \, a b^{5} c^{4} - 48 \, a^{3} b^{3} c^{2} d^{2} + 80 \, a^{4} b^{2} c d^{3} - 35 \, a^{5} b d^{4}\right )} x\right )} \sqrt{b x^{2} + a}}{48 \,{\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, -\frac{3 \,{\left (48 \, a^{4} b^{2} c^{2} d^{2} - 80 \, a^{5} b c d^{3} + 35 \, a^{6} d^{4} +{\left (48 \, a^{2} b^{4} c^{2} d^{2} - 80 \, a^{3} b^{3} c d^{3} + 35 \, a^{4} b^{2} d^{4}\right )} x^{4} + 2 \,{\left (48 \, a^{3} b^{3} c^{2} d^{2} - 80 \, a^{4} b^{2} c d^{3} + 35 \, a^{5} b d^{4}\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (6 \, a^{2} b^{4} d^{4} x^{7} + 3 \,{\left (16 \, a^{2} b^{4} c d^{3} - 7 \, a^{3} b^{3} d^{4}\right )} x^{5} + 4 \,{\left (4 \, b^{6} c^{4} + 8 \, a b^{5} c^{3} d - 48 \, a^{2} b^{4} c^{2} d^{2} + 80 \, a^{3} b^{3} c d^{3} - 35 \, a^{4} b^{2} d^{4}\right )} x^{3} + 3 \,{\left (8 \, a b^{5} c^{4} - 48 \, a^{3} b^{3} c^{2} d^{2} + 80 \, a^{4} b^{2} c d^{3} - 35 \, a^{5} b d^{4}\right )} x\right )} \sqrt{b x^{2} + a}}{24 \,{\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*(48*a^4*b^2*c^2*d^2 - 80*a^5*b*c*d^3 + 35*a^6*d^4 + (48*a^2*b^4*c^2*d^2 - 80*a^3*b^3*c*d^3 + 35*a^4*b
^2*d^4)*x^4 + 2*(48*a^3*b^3*c^2*d^2 - 80*a^4*b^2*c*d^3 + 35*a^5*b*d^4)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^
2 + a)*sqrt(b)*x - a) + 2*(6*a^2*b^4*d^4*x^7 + 3*(16*a^2*b^4*c*d^3 - 7*a^3*b^3*d^4)*x^5 + 4*(4*b^6*c^4 + 8*a*b
^5*c^3*d - 48*a^2*b^4*c^2*d^2 + 80*a^3*b^3*c*d^3 - 35*a^4*b^2*d^4)*x^3 + 3*(8*a*b^5*c^4 - 48*a^3*b^3*c^2*d^2 +
80*a^4*b^2*c*d^3 - 35*a^5*b*d^4)*x)*sqrt(b*x^2 + a))/(a^2*b^7*x^4 + 2*a^3*b^6*x^2 + a^4*b^5), -1/24*(3*(48*a^
4*b^2*c^2*d^2 - 80*a^5*b*c*d^3 + 35*a^6*d^4 + (48*a^2*b^4*c^2*d^2 - 80*a^3*b^3*c*d^3 + 35*a^4*b^2*d^4)*x^4 + 2
*(48*a^3*b^3*c^2*d^2 - 80*a^4*b^2*c*d^3 + 35*a^5*b*d^4)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (6*
a^2*b^4*d^4*x^7 + 3*(16*a^2*b^4*c*d^3 - 7*a^3*b^3*d^4)*x^5 + 4*(4*b^6*c^4 + 8*a*b^5*c^3*d - 48*a^2*b^4*c^2*d^2
+ 80*a^3*b^3*c*d^3 - 35*a^4*b^2*d^4)*x^3 + 3*(8*a*b^5*c^4 - 48*a^3*b^3*c^2*d^2 + 80*a^4*b^2*c*d^3 - 35*a^5*b*
d^4)*x)*sqrt(b*x^2 + a))/(a^2*b^7*x^4 + 2*a^3*b^6*x^2 + a^4*b^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{2}\right )^{4}}{\left (a + b x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18895, size = 320, normalized size = 1.25 \begin{align*} \frac{{\left ({\left (3 \,{\left (\frac{2 \, d^{4} x^{2}}{b} + \frac{16 \, a^{2} b^{6} c d^{3} - 7 \, a^{3} b^{5} d^{4}}{a^{2} b^{7}}\right )} x^{2} + \frac{4 \,{\left (4 \, b^{8} c^{4} + 8 \, a b^{7} c^{3} d - 48 \, a^{2} b^{6} c^{2} d^{2} + 80 \, a^{3} b^{5} c d^{3} - 35 \, a^{4} b^{4} d^{4}\right )}}{a^{2} b^{7}}\right )} x^{2} + \frac{3 \,{\left (8 \, a b^{7} c^{4} - 48 \, a^{3} b^{5} c^{2} d^{2} + 80 \, a^{4} b^{4} c d^{3} - 35 \, a^{5} b^{3} d^{4}\right )}}{a^{2} b^{7}}\right )} x}{24 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} - \frac{{\left (48 \, b^{2} c^{2} d^{2} - 80 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{8 \, b^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((3*(2*d^4*x^2/b + (16*a^2*b^6*c*d^3 - 7*a^3*b^5*d^4)/(a^2*b^7))*x^2 + 4*(4*b^8*c^4 + 8*a*b^7*c^3*d - 48*
a^2*b^6*c^2*d^2 + 80*a^3*b^5*c*d^3 - 35*a^4*b^4*d^4)/(a^2*b^7))*x^2 + 3*(8*a*b^7*c^4 - 48*a^3*b^5*c^2*d^2 + 80
*a^4*b^4*c*d^3 - 35*a^5*b^3*d^4)/(a^2*b^7))*x/(b*x^2 + a)^(3/2) - 1/8*(48*b^2*c^2*d^2 - 80*a*b*c*d^3 + 35*a^2*
d^4)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(9/2)