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

Optimal. Leaf size=169 \[ \frac {(2 b c-a d) \left (5 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}}+\frac {d x \sqrt {a+b x^2} \left (15 a^2 d^2-44 a b c d+44 b^2 c^2\right )}{48 b^3}+\frac {5 d x \sqrt {a+b x^2} \left (c+d x^2\right ) (2 b c-a d)}{24 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b} \]

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Rubi [A]  time = 0.14, 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 = {416, 528, 388, 217, 206} \begin {gather*} \frac {d x \sqrt {a+b x^2} \left (15 a^2 d^2-44 a b c d+44 b^2 c^2\right )}{48 b^3}+\frac {(2 b c-a d) \left (5 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}}+\frac {5 d x \sqrt {a+b x^2} \left (c+d x^2\right ) (2 b c-a d)}{24 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^3/Sqrt[a + b*x^2],x]

[Out]

(d*(44*b^2*c^2 - 44*a*b*c*d + 15*a^2*d^2)*x*Sqrt[a + b*x^2])/(48*b^3) + (5*d*(2*b*c - a*d)*x*Sqrt[a + b*x^2]*(
c + d*x^2))/(24*b^2) + (d*x*Sqrt[a + b*x^2]*(c + d*x^2)^2)/(6*b) + ((2*b*c - a*d)*(8*b^2*c^2 - 8*a*b*c*d + 5*a
^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*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 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[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}{\sqrt {a+b x^2}} \, dx &=\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}+\frac {\int \frac {\left (c+d x^2\right ) \left (c (6 b c-a d)+5 d (2 b c-a d) x^2\right )}{\sqrt {a+b x^2}} \, dx}{6 b}\\ &=\frac {5 d (2 b c-a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}+\frac {\int \frac {c \left (24 b^2 c^2-14 a b c d+5 a^2 d^2\right )+d \left (44 b^2 c^2-44 a b c d+15 a^2 d^2\right ) x^2}{\sqrt {a+b x^2}} \, dx}{24 b^2}\\ &=\frac {d \left (44 b^2 c^2-44 a b c d+15 a^2 d^2\right ) x \sqrt {a+b x^2}}{48 b^3}+\frac {5 d (2 b c-a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}+\frac {\left ((2 b c-a d) \left (8 b^2 c^2-8 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^3}\\ &=\frac {d \left (44 b^2 c^2-44 a b c d+15 a^2 d^2\right ) x \sqrt {a+b x^2}}{48 b^3}+\frac {5 d (2 b c-a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}+\frac {\left ((2 b c-a d) \left (8 b^2 c^2-8 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 )}{16 b^3}\\ &=\frac {d \left (44 b^2 c^2-44 a b c d+15 a^2 d^2\right ) x \sqrt {a+b x^2}}{48 b^3}+\frac {5 d (2 b c-a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{24 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}+\frac {(2 b c-a d) \left (8 b^2 c^2-8 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 5.09, size = 140, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b} d x \sqrt {a+b x^2} \left (15 a^2 d^2-2 a b d \left (27 c+5 d x^2\right )+4 b^2 \left (18 c^2+9 c d x^2+2 d^2 x^4\right )\right )+3 \left (-5 a^3 d^3+18 a^2 b c d^2-24 a b^2 c^2 d+16 b^3 c^3\right ) \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{48 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*d*x*Sqrt[a + b*x^2]*(15*a^2*d^2 - 2*a*b*d*(27*c + 5*d*x^2) + 4*b^2*(18*c^2 + 9*c*d*x^2 + 2*d^2*x^4))
+ 3*(16*b^3*c^3 - 24*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 5*a^3*d^3)*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(48*b^(7/2)
)

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IntegrateAlgebraic [A]  time = 0.17, size = 148, normalized size = 0.88 \begin {gather*} \frac {\sqrt {a+b x^2} \left (15 a^2 d^3 x-54 a b c d^2 x-10 a b d^3 x^3+72 b^2 c^2 d x+36 b^2 c d^2 x^3+8 b^2 d^3 x^5\right )}{48 b^3}+\frac {\left (5 a^3 d^3-18 a^2 b c d^2+24 a b^2 c^2 d-16 b^3 c^3\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{16 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(72*b^2*c^2*d*x - 54*a*b*c*d^2*x + 15*a^2*d^3*x + 36*b^2*c*d^2*x^3 - 10*a*b*d^3*x^3 + 8*b^2*d
^3*x^5))/(48*b^3) + ((-16*b^3*c^3 + 24*a*b^2*c^2*d - 18*a^2*b*c*d^2 + 5*a^3*d^3)*Log[-(Sqrt[b]*x) + Sqrt[a + b
*x^2]])/(16*b^(7/2))

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fricas [A]  time = 1.18, size = 300, normalized size = 1.78 \begin {gather*} \left [-\frac {3 \, {\left (16 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{5} + 2 \, {\left (18 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{3} + 3 \, {\left (24 \, b^{3} c^{2} d - 18 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, -\frac {3 \, {\left (16 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d^{3} x^{5} + 2 \, {\left (18 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{3} + 3 \, {\left (24 \, b^{3} c^{2} d - 18 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(16*b^3*c^3 - 24*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*
sqrt(b)*x - a) - 2*(8*b^3*d^3*x^5 + 2*(18*b^3*c*d^2 - 5*a*b^2*d^3)*x^3 + 3*(24*b^3*c^2*d - 18*a*b^2*c*d^2 + 5*
a^2*b*d^3)*x)*sqrt(b*x^2 + a))/b^4, -1/48*(3*(16*b^3*c^3 - 24*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-
b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*d^3*x^5 + 2*(18*b^3*c*d^2 - 5*a*b^2*d^3)*x^3 + 3*(24*b^3*c^2*d
- 18*a*b^2*c*d^2 + 5*a^2*b*d^3)*x)*sqrt(b*x^2 + a))/b^4]

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giac [A]  time = 0.64, size = 150, normalized size = 0.89 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, d^{3} x^{2}}{b} + \frac {18 \, b^{4} c d^{2} - 5 \, a b^{3} d^{3}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (24 \, b^{4} c^{2} d - 18 \, a b^{3} c d^{2} + 5 \, a^{2} b^{2} d^{3}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (16 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, 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)^(1/2),x, algorithm="giac")

[Out]

1/48*(2*(4*d^3*x^2/b + (18*b^4*c*d^2 - 5*a*b^3*d^3)/b^5)*x^2 + 3*(24*b^4*c^2*d - 18*a*b^3*c*d^2 + 5*a^2*b^2*d^
3)/b^5)*sqrt(b*x^2 + a)*x - 1/16*(16*b^3*c^3 - 24*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 5*a^3*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 = 228, normalized size = 1.35 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, d^{3} x^{5}}{6 b}-\frac {5 \sqrt {b \,x^{2}+a}\, a \,d^{3} x^{3}}{24 b^{2}}+\frac {3 \sqrt {b \,x^{2}+a}\, c \,d^{2} x^{3}}{4 b}-\frac {5 a^{3} d^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}+\frac {9 a^{2} c \,d^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}-\frac {3 a \,c^{2} d \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {c^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {5 \sqrt {b \,x^{2}+a}\, a^{2} d^{3} x}{16 b^{3}}-\frac {9 \sqrt {b \,x^{2}+a}\, a c \,d^{2} x}{8 b^{2}}+\frac {3 \sqrt {b \,x^{2}+a}\, c^{2} d x}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^3/(b*x^2+a)^(1/2),x)

[Out]

1/6*d^3*x^5/b*(b*x^2+a)^(1/2)-5/24*d^3*a/b^2*x^3*(b*x^2+a)^(1/2)+5/16*d^3*a^2/b^3*x*(b*x^2+a)^(1/2)-5/16*d^3*a
^3/b^(7/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+3/4*c*d^2*x^3/b*(b*x^2+a)^(1/2)-9/8*c*d^2*a/b^2*x*(b*x^2+a)^(1/2)+9/8
*c*d^2*a^2/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+3/2*c^2*d*x/b*(b*x^2+a)^(1/2)-3/2*c^2*d*a/b^(3/2)*ln(b^(1/2)*
x+(b*x^2+a)^(1/2))+c^3*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)

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maxima [A]  time = 1.36, size = 199, normalized size = 1.18 \begin {gather*} \frac {\sqrt {b x^{2} + a} d^{3} x^{5}}{6 \, b} + \frac {3 \, \sqrt {b x^{2} + a} c d^{2} x^{3}}{4 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a d^{3} x^{3}}{24 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} c^{2} d x}{2 \, b} - \frac {9 \, \sqrt {b x^{2} + a} a c d^{2} x}{8 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a} a^{2} d^{3} x}{16 \, b^{3}} + \frac {c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {3 \, a c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {9 \, a^{2} c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {5 \, a^{3} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, 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)^(1/2),x, algorithm="maxima")

[Out]

1/6*sqrt(b*x^2 + a)*d^3*x^5/b + 3/4*sqrt(b*x^2 + a)*c*d^2*x^3/b - 5/24*sqrt(b*x^2 + a)*a*d^3*x^3/b^2 + 3/2*sqr
t(b*x^2 + a)*c^2*d*x/b - 9/8*sqrt(b*x^2 + a)*a*c*d^2*x/b^2 + 5/16*sqrt(b*x^2 + a)*a^2*d^3*x/b^3 + c^3*arcsinh(
b*x/sqrt(a*b))/sqrt(b) - 3/2*a*c^2*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 9/8*a^2*c*d^2*arcsinh(b*x/sqrt(a*b))/b^(
5/2) - 5/16*a^3*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}{\sqrt {b\,x^2+a}} \,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)^(1/2),x)

[Out]

int((c + d*x^2)^3/(a + b*x^2)^(1/2), x)

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sympy [A]  time = 13.27, size = 400, normalized size = 2.37 \begin {gather*} \frac {5 a^{\frac {5}{2}} d^{3} x}{16 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {9 a^{\frac {3}{2}} c d^{2} x}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 a^{\frac {3}{2}} d^{3} x^{3}}{48 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 \sqrt {a} c^{2} d x \sqrt {1 + \frac {b x^{2}}{a}}}{2 b} - \frac {3 \sqrt {a} c d^{2} x^{3}}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {\sqrt {a} d^{3} x^{5}}{24 b \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 a^{3} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {7}{2}}} + \frac {9 a^{2} c d^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} - \frac {3 a c^{2} d \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} + c^{3} \left (\begin {cases} \frac {\sqrt {- \frac {a}{b}} \operatorname {asin}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b < 0 \\\frac {\sqrt {\frac {a}{b}} \operatorname {asinh}{\left (x \sqrt {\frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b > 0 \\\frac {\sqrt {- \frac {a}{b}} \operatorname {acosh}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {- a}} & \text {for}\: b > 0 \wedge a < 0 \end {cases}\right ) + \frac {3 c d^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {d^{3} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**3/(b*x**2+a)**(1/2),x)

[Out]

5*a**(5/2)*d**3*x/(16*b**3*sqrt(1 + b*x**2/a)) - 9*a**(3/2)*c*d**2*x/(8*b**2*sqrt(1 + b*x**2/a)) + 5*a**(3/2)*
d**3*x**3/(48*b**2*sqrt(1 + b*x**2/a)) + 3*sqrt(a)*c**2*d*x*sqrt(1 + b*x**2/a)/(2*b) - 3*sqrt(a)*c*d**2*x**3/(
8*b*sqrt(1 + b*x**2/a)) - sqrt(a)*d**3*x**5/(24*b*sqrt(1 + b*x**2/a)) - 5*a**3*d**3*asinh(sqrt(b)*x/sqrt(a))/(
16*b**(7/2)) + 9*a**2*c*d**2*asinh(sqrt(b)*x/sqrt(a))/(8*b**(5/2)) - 3*a*c**2*d*asinh(sqrt(b)*x/sqrt(a))/(2*b*
*(3/2)) + c**3*Piecewise((sqrt(-a/b)*asin(x*sqrt(-b/a))/sqrt(a), (a > 0) & (b < 0)), (sqrt(a/b)*asinh(x*sqrt(b
/a))/sqrt(a), (a > 0) & (b > 0)), (sqrt(-a/b)*acosh(x*sqrt(-b/a))/sqrt(-a), (b > 0) & (a < 0))) + 3*c*d**2*x**
5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + d**3*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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