3.1.81 \(\int \frac {(c+d x^2)^4}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=257 \[ -\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right ) \left (35 a^2 d^2-64 a b c d+24 b^2 c^2\right )}{24 a b^3}+\frac {d \left (-35 a^3 d^3+120 a^2 b c d^2-144 a b^2 c^2 d+64 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}}-\frac {d x \sqrt {a+b x^2} \left (-105 a^3 d^3+290 a^2 b c d^2-248 a b^2 c^2 d+48 b^3 c^3\right )}{48 a b^4}-\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2 (6 b c-7 a d)}{6 a b^2}+\frac {x \left (c+d x^2\right )^3 (b c-a d)}{a b \sqrt {a+b x^2}} \]

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Rubi [A]  time = 0.26, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {d x \sqrt {a+b x^2} \left (c+d x^2\right ) \left (35 a^2 d^2-64 a b c d+24 b^2 c^2\right )}{24 a b^3}-\frac {d x \sqrt {a+b x^2} \left (290 a^2 b c d^2-105 a^3 d^3-248 a b^2 c^2 d+48 b^3 c^3\right )}{48 a b^4}+\frac {d \left (120 a^2 b c d^2-35 a^3 d^3-144 a b^2 c^2 d+64 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}}-\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2 (6 b c-7 a d)}{6 a b^2}+\frac {x \left (c+d x^2\right )^3 (b c-a d)}{a b \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^4/(a + b*x^2)^(3/2),x]

[Out]

-(d*(48*b^3*c^3 - 248*a*b^2*c^2*d + 290*a^2*b*c*d^2 - 105*a^3*d^3)*x*Sqrt[a + b*x^2])/(48*a*b^4) - (d*(24*b^2*
c^2 - 64*a*b*c*d + 35*a^2*d^2)*x*Sqrt[a + b*x^2]*(c + d*x^2))/(24*a*b^3) - (d*(6*b*c - 7*a*d)*x*Sqrt[a + b*x^2
]*(c + d*x^2)^2)/(6*a*b^2) + ((b*c - a*d)*x*(c + d*x^2)^3)/(a*b*Sqrt[a + b*x^2]) + (d*(64*b^3*c^3 - 144*a*b^2*
c^2*d + 120*a^2*b*c*d^2 - 35*a^3*d^3)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(9/2))

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 )^4}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac {(b c-a d) x \left (c+d x^2\right )^3}{a b \sqrt {a+b x^2}}+\frac {\int \frac {\left (c+d x^2\right )^2 \left (a c d-d (6 b c-7 a d) x^2\right )}{\sqrt {a+b x^2}} \, dx}{a b}\\ &=-\frac {d (6 b c-7 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^3}{a b \sqrt {a+b x^2}}+\frac {\int \frac {\left (c+d x^2\right ) \left (a c d (12 b c-7 a d)-d \left (24 b^2 c^2-64 a b c d+35 a^2 d^2\right ) x^2\right )}{\sqrt {a+b x^2}} \, dx}{6 a b^2}\\ &=-\frac {d \left (24 b^2 c^2-64 a b c d+35 a^2 d^2\right ) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 a b^3}-\frac {d (6 b c-7 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^3}{a b \sqrt {a+b x^2}}+\frac {\int \frac {a c d \left (72 b^2 c^2-92 a b c d+35 a^2 d^2\right )-d \left (48 b^3 c^3-248 a b^2 c^2 d+290 a^2 b c d^2-105 a^3 d^3\right ) x^2}{\sqrt {a+b x^2}} \, dx}{24 a b^3}\\ &=-\frac {d \left (48 b^3 c^3-248 a b^2 c^2 d+290 a^2 b c d^2-105 a^3 d^3\right ) x \sqrt {a+b x^2}}{48 a b^4}-\frac {d \left (24 b^2 c^2-64 a b c d+35 a^2 d^2\right ) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 a b^3}-\frac {d (6 b c-7 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^3}{a b \sqrt {a+b x^2}}+\frac {\left (d \left (64 b^3 c^3-144 a b^2 c^2 d+120 a^2 b c d^2-35 a^3 d^3\right )\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^4}\\ &=-\frac {d \left (48 b^3 c^3-248 a b^2 c^2 d+290 a^2 b c d^2-105 a^3 d^3\right ) x \sqrt {a+b x^2}}{48 a b^4}-\frac {d \left (24 b^2 c^2-64 a b c d+35 a^2 d^2\right ) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 a b^3}-\frac {d (6 b c-7 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^3}{a b \sqrt {a+b x^2}}+\frac {\left (d \left (64 b^3 c^3-144 a b^2 c^2 d+120 a^2 b c d^2-35 a^3 d^3\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 b^4}\\ &=-\frac {d \left (48 b^3 c^3-248 a b^2 c^2 d+290 a^2 b c d^2-105 a^3 d^3\right ) x \sqrt {a+b x^2}}{48 a b^4}-\frac {d \left (24 b^2 c^2-64 a b c d+35 a^2 d^2\right ) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 a b^3}-\frac {d (6 b c-7 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^3}{a b \sqrt {a+b x^2}}+\frac {d \left (64 b^3 c^3-144 a b^2 c^2 d+120 a^2 b c d^2-35 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 5.20, size = 172, normalized size = 0.67 \begin {gather*} \frac {\sqrt {b} x \sqrt {a+b x^2} \left (3 d^2 \left (19 a^2 d^2-56 a b c d+48 b^2 c^2\right )+2 b d^3 x^2 (24 b c-11 a d)+\frac {48 (b c-a d)^4}{a \left (a+b x^2\right )}+8 b^2 d^4 x^4\right )+3 d \left (-35 a^3 d^3+120 a^2 b c d^2-144 a b^2 c^2 d+64 b^3 c^3\right ) \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{48 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^4/(a + b*x^2)^(3/2),x]

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(3*d^2*(48*b^2*c^2 - 56*a*b*c*d + 19*a^2*d^2) + 2*b*d^3*(24*b*c - 11*a*d)*x^2 + 8*b
^2*d^4*x^4 + (48*(b*c - a*d)^4)/(a*(a + b*x^2))) + 3*d*(64*b^3*c^3 - 144*a*b^2*c^2*d + 120*a^2*b*c*d^2 - 35*a^
3*d^3)*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(48*b^(9/2))

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IntegrateAlgebraic [A]  time = 0.48, size = 229, normalized size = 0.89 \begin {gather*} \frac {\left (35 a^3 d^4-120 a^2 b c d^3+144 a b^2 c^2 d^2-64 b^3 c^3 d\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{16 b^{9/2}}+\frac {105 a^4 d^4 x-360 a^3 b c d^3 x+35 a^3 b d^4 x^3+432 a^2 b^2 c^2 d^2 x-120 a^2 b^2 c d^3 x^3-14 a^2 b^2 d^4 x^5-192 a b^3 c^3 d x+144 a b^3 c^2 d^2 x^3+48 a b^3 c d^3 x^5+8 a b^3 d^4 x^7+48 b^4 c^4 x}{48 a b^4 \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2)^4/(a + b*x^2)^(3/2),x]

[Out]

(48*b^4*c^4*x - 192*a*b^3*c^3*d*x + 432*a^2*b^2*c^2*d^2*x - 360*a^3*b*c*d^3*x + 105*a^4*d^4*x + 144*a*b^3*c^2*
d^2*x^3 - 120*a^2*b^2*c*d^3*x^3 + 35*a^3*b*d^4*x^3 + 48*a*b^3*c*d^3*x^5 - 14*a^2*b^2*d^4*x^5 + 8*a*b^3*d^4*x^7
)/(48*a*b^4*Sqrt[a + b*x^2]) + ((-64*b^3*c^3*d + 144*a*b^2*c^2*d^2 - 120*a^2*b*c*d^3 + 35*a^3*d^4)*Log[-(Sqrt[
b]*x) + Sqrt[a + b*x^2]])/(16*b^(9/2))

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fricas [A]  time = 1.39, size = 584, normalized size = 2.27 \begin {gather*} \left [-\frac {3 \, {\left (64 \, a^{2} b^{3} c^{3} d - 144 \, a^{3} b^{2} c^{2} d^{2} + 120 \, a^{4} b c d^{3} - 35 \, a^{5} d^{4} + {\left (64 \, a b^{4} c^{3} d - 144 \, a^{2} b^{3} c^{2} d^{2} + 120 \, a^{3} b^{2} c d^{3} - 35 \, a^{4} 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 (8 \, a b^{4} d^{4} x^{7} + 2 \, {\left (24 \, a b^{4} c d^{3} - 7 \, a^{2} b^{3} d^{4}\right )} x^{5} + {\left (144 \, a b^{4} c^{2} d^{2} - 120 \, a^{2} b^{3} c d^{3} + 35 \, a^{3} b^{2} d^{4}\right )} x^{3} + 3 \, {\left (16 \, b^{5} c^{4} - 64 \, a b^{4} c^{3} d + 144 \, a^{2} b^{3} c^{2} d^{2} - 120 \, a^{3} b^{2} c d^{3} + 35 \, a^{4} b d^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, {\left (a b^{6} x^{2} + a^{2} b^{5}\right )}}, -\frac {3 \, {\left (64 \, a^{2} b^{3} c^{3} d - 144 \, a^{3} b^{2} c^{2} d^{2} + 120 \, a^{4} b c d^{3} - 35 \, a^{5} d^{4} + {\left (64 \, a b^{4} c^{3} d - 144 \, a^{2} b^{3} c^{2} d^{2} + 120 \, a^{3} b^{2} c d^{3} - 35 \, a^{4} b d^{4}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, a b^{4} d^{4} x^{7} + 2 \, {\left (24 \, a b^{4} c d^{3} - 7 \, a^{2} b^{3} d^{4}\right )} x^{5} + {\left (144 \, a b^{4} c^{2} d^{2} - 120 \, a^{2} b^{3} c d^{3} + 35 \, a^{3} b^{2} d^{4}\right )} x^{3} + 3 \, {\left (16 \, b^{5} c^{4} - 64 \, a b^{4} c^{3} d + 144 \, a^{2} b^{3} c^{2} d^{2} - 120 \, a^{3} b^{2} c d^{3} + 35 \, a^{4} b d^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, {\left (a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^4/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(64*a^2*b^3*c^3*d - 144*a^3*b^2*c^2*d^2 + 120*a^4*b*c*d^3 - 35*a^5*d^4 + (64*a*b^4*c^3*d - 144*a^2*b
^3*c^2*d^2 + 120*a^3*b^2*c*d^3 - 35*a^4*b*d^4)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) -
2*(8*a*b^4*d^4*x^7 + 2*(24*a*b^4*c*d^3 - 7*a^2*b^3*d^4)*x^5 + (144*a*b^4*c^2*d^2 - 120*a^2*b^3*c*d^3 + 35*a^3*
b^2*d^4)*x^3 + 3*(16*b^5*c^4 - 64*a*b^4*c^3*d + 144*a^2*b^3*c^2*d^2 - 120*a^3*b^2*c*d^3 + 35*a^4*b*d^4)*x)*sqr
t(b*x^2 + a))/(a*b^6*x^2 + a^2*b^5), -1/48*(3*(64*a^2*b^3*c^3*d - 144*a^3*b^2*c^2*d^2 + 120*a^4*b*c*d^3 - 35*a
^5*d^4 + (64*a*b^4*c^3*d - 144*a^2*b^3*c^2*d^2 + 120*a^3*b^2*c*d^3 - 35*a^4*b*d^4)*x^2)*sqrt(-b)*arctan(sqrt(-
b)*x/sqrt(b*x^2 + a)) - (8*a*b^4*d^4*x^7 + 2*(24*a*b^4*c*d^3 - 7*a^2*b^3*d^4)*x^5 + (144*a*b^4*c^2*d^2 - 120*a
^2*b^3*c*d^3 + 35*a^3*b^2*d^4)*x^3 + 3*(16*b^5*c^4 - 64*a*b^4*c^3*d + 144*a^2*b^3*c^2*d^2 - 120*a^3*b^2*c*d^3
+ 35*a^4*b*d^4)*x)*sqrt(b*x^2 + a))/(a*b^6*x^2 + a^2*b^5)]

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giac [A]  time = 0.67, size = 235, normalized size = 0.91 \begin {gather*} \frac {{\left ({\left (2 \, {\left (\frac {4 \, d^{4} x^{2}}{b} + \frac {24 \, a b^{6} c d^{3} - 7 \, a^{2} b^{5} d^{4}}{a b^{7}}\right )} x^{2} + \frac {144 \, a b^{6} c^{2} d^{2} - 120 \, a^{2} b^{5} c d^{3} + 35 \, a^{3} b^{4} d^{4}}{a b^{7}}\right )} x^{2} + \frac {3 \, {\left (16 \, b^{7} c^{4} - 64 \, a b^{6} c^{3} d + 144 \, a^{2} b^{5} c^{2} d^{2} - 120 \, a^{3} b^{4} c d^{3} + 35 \, a^{4} b^{3} d^{4}\right )}}{a b^{7}}\right )} x}{48 \, \sqrt {b x^{2} + a}} - \frac {{\left (64 \, b^{3} c^{3} d - 144 \, a b^{2} c^{2} d^{2} + 120 \, a^{2} b c d^{3} - 35 \, a^{3} d^{4}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^4/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/48*((2*(4*d^4*x^2/b + (24*a*b^6*c*d^3 - 7*a^2*b^5*d^4)/(a*b^7))*x^2 + (144*a*b^6*c^2*d^2 - 120*a^2*b^5*c*d^3
+ 35*a^3*b^4*d^4)/(a*b^7))*x^2 + 3*(16*b^7*c^4 - 64*a*b^6*c^3*d + 144*a^2*b^5*c^2*d^2 - 120*a^3*b^4*c*d^3 + 3
5*a^4*b^3*d^4)/(a*b^7))*x/sqrt(b*x^2 + a) - 1/16*(64*b^3*c^3*d - 144*a*b^2*c^2*d^2 + 120*a^2*b*c*d^3 - 35*a^3*
d^4)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(9/2)

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maple [A]  time = 0.02, size = 340, normalized size = 1.32 \begin {gather*} \frac {d^{4} x^{7}}{6 \sqrt {b \,x^{2}+a}\, b}-\frac {7 a \,d^{4} x^{5}}{24 \sqrt {b \,x^{2}+a}\, b^{2}}+\frac {c \,d^{3} x^{5}}{\sqrt {b \,x^{2}+a}\, b}+\frac {35 a^{2} d^{4} x^{3}}{48 \sqrt {b \,x^{2}+a}\, b^{3}}-\frac {5 a c \,d^{3} x^{3}}{2 \sqrt {b \,x^{2}+a}\, b^{2}}+\frac {3 c^{2} d^{2} x^{3}}{\sqrt {b \,x^{2}+a}\, b}+\frac {35 a^{3} d^{4} x}{16 \sqrt {b \,x^{2}+a}\, b^{4}}-\frac {15 a^{2} c \,d^{3} x}{2 \sqrt {b \,x^{2}+a}\, b^{3}}+\frac {9 a \,c^{2} d^{2} x}{\sqrt {b \,x^{2}+a}\, b^{2}}+\frac {c^{4} x}{\sqrt {b \,x^{2}+a}\, a}-\frac {4 c^{3} d x}{\sqrt {b \,x^{2}+a}\, b}-\frac {35 a^{3} d^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {9}{2}}}+\frac {15 a^{2} c \,d^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {7}{2}}}-\frac {9 a \,c^{2} d^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {5}{2}}}+\frac {4 c^{3} 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)^4/(b*x^2+a)^(3/2),x)

[Out]

1/6*d^4*x^7/b/(b*x^2+a)^(1/2)-7/24*d^4*a/b^2*x^5/(b*x^2+a)^(1/2)+35/48*d^4*a^2/b^3*x^3/(b*x^2+a)^(1/2)+35/16*d
^4*a^3/b^4*x/(b*x^2+a)^(1/2)-35/16*d^4*a^3/b^(9/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+c*d^3*x^5/b/(b*x^2+a)^(1/2)-5
/2*c*d^3*a/b^2*x^3/(b*x^2+a)^(1/2)-15/2*c*d^3*a^2/b^3*x/(b*x^2+a)^(1/2)+15/2*c*d^3*a^2/b^(7/2)*ln(b^(1/2)*x+(b
*x^2+a)^(1/2))+3*c^2*d^2*x^3/b/(b*x^2+a)^(1/2)+9*c^2*d^2*a/b^2*x/(b*x^2+a)^(1/2)-9*c^2*d^2*a/b^(5/2)*ln(b^(1/2
)*x+(b*x^2+a)^(1/2))-4*c^3*d*x/b/(b*x^2+a)^(1/2)+4*c^3*d/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+c^4*x/a/(b*x^2+
a)^(1/2)

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maxima [A]  time = 1.38, size = 311, normalized size = 1.21 \begin {gather*} \frac {d^{4} x^{7}}{6 \, \sqrt {b x^{2} + a} b} + \frac {c d^{3} x^{5}}{\sqrt {b x^{2} + a} b} - \frac {7 \, a d^{4} x^{5}}{24 \, \sqrt {b x^{2} + a} b^{2}} + \frac {3 \, c^{2} d^{2} x^{3}}{\sqrt {b x^{2} + a} b} - \frac {5 \, a c d^{3} x^{3}}{2 \, \sqrt {b x^{2} + a} b^{2}} + \frac {35 \, a^{2} d^{4} x^{3}}{48 \, \sqrt {b x^{2} + a} b^{3}} + \frac {c^{4} x}{\sqrt {b x^{2} + a} a} - \frac {4 \, c^{3} d x}{\sqrt {b x^{2} + a} b} + \frac {9 \, a c^{2} d^{2} x}{\sqrt {b x^{2} + a} b^{2}} - \frac {15 \, a^{2} c d^{3} x}{2 \, \sqrt {b x^{2} + a} b^{3}} + \frac {35 \, a^{3} d^{4} x}{16 \, \sqrt {b x^{2} + a} b^{4}} + \frac {4 \, c^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {9 \, a c^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} + \frac {15 \, a^{2} c d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} - \frac {35 \, a^{3} d^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^4/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

1/6*d^4*x^7/(sqrt(b*x^2 + a)*b) + c*d^3*x^5/(sqrt(b*x^2 + a)*b) - 7/24*a*d^4*x^5/(sqrt(b*x^2 + a)*b^2) + 3*c^2
*d^2*x^3/(sqrt(b*x^2 + a)*b) - 5/2*a*c*d^3*x^3/(sqrt(b*x^2 + a)*b^2) + 35/48*a^2*d^4*x^3/(sqrt(b*x^2 + a)*b^3)
+ c^4*x/(sqrt(b*x^2 + a)*a) - 4*c^3*d*x/(sqrt(b*x^2 + a)*b) + 9*a*c^2*d^2*x/(sqrt(b*x^2 + a)*b^2) - 15/2*a^2*
c*d^3*x/(sqrt(b*x^2 + a)*b^3) + 35/16*a^3*d^4*x/(sqrt(b*x^2 + a)*b^4) + 4*c^3*d*arcsinh(b*x/sqrt(a*b))/b^(3/2)
- 9*a*c^2*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 15/2*a^2*c*d^3*arcsinh(b*x/sqrt(a*b))/b^(7/2) - 35/16*a^3*d^4*
arcsinh(b*x/sqrt(a*b))/b^(9/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^4}{{\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)^4/(a + b*x^2)^(3/2),x)

[Out]

int((c + d*x^2)^4/(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 )^{4}}{\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)**4/(b*x**2+a)**(3/2),x)

[Out]

Integral((c + d*x**2)**4/(a + b*x**2)**(3/2), x)

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