3.12.42 \(\int \frac {(-q+p x^2) (a q+b x+a p x^2) \sqrt {q^2+p^2 x^4}}{x^4} \, dx\)

Optimal. Leaf size=85 \[ \frac {\sqrt {p^2 x^4+q^2} \left (2 a p^2 x^4+2 a q^2+3 b p x^3+3 b q x\right )}{6 x^3}-b p q \log \left (\sqrt {p^2 x^4+q^2}+p x^2+q\right )+b p q \log (x) \]

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Rubi [A]  time = 0.24, antiderivative size = 109, normalized size of antiderivative = 1.28, number of steps used = 11, number of rules used = 10, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {1833, 1252, 813, 844, 217, 206, 266, 63, 208, 449} \begin {gather*} \frac {a \left (p^2 x^4+q^2\right )^{3/2}}{3 x^3}-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {\sqrt {p^2 x^4+q^2}}{q}\right )+\frac {b \left (p x^2+q\right ) \sqrt {p^2 x^4+q^2}}{2 x^2}-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {p x^2}{\sqrt {p^2 x^4+q^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-q + p*x^2)*(a*q + b*x + a*p*x^2)*Sqrt[q^2 + p^2*x^4])/x^4,x]

[Out]

(b*(q + p*x^2)*Sqrt[q^2 + p^2*x^4])/(2*x^2) + (a*(q^2 + p^2*x^4)^(3/2))/(3*x^3) - (b*p*q*ArcTanh[(p*x^2)/Sqrt[
q^2 + p^2*x^4]])/2 - (b*p*q*ArcTanh[Sqrt[q^2 + p^2*x^4]/q])/2

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 449

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

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 1833

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

\begin {align*} \int \frac {\left (-q+p x^2\right ) \left (a q+b x+a p x^2\right ) \sqrt {q^2+p^2 x^4}}{x^4} \, dx &=\int \left (\frac {\left (-b q+b p x^2\right ) \sqrt {q^2+p^2 x^4}}{x^3}+\frac {\sqrt {q^2+p^2 x^4} \left (-a q^2+a p^2 x^4\right )}{x^4}\right ) \, dx\\ &=\int \frac {\left (-b q+b p x^2\right ) \sqrt {q^2+p^2 x^4}}{x^3} \, dx+\int \frac {\sqrt {q^2+p^2 x^4} \left (-a q^2+a p^2 x^4\right )}{x^4} \, dx\\ &=\frac {a \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {(-b q+b p x) \sqrt {q^2+p^2 x^2}}{x^2} \, dx,x,x^2\right )\\ &=\frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {-2 b p q^2+2 b p^2 q x}{x \sqrt {q^2+p^2 x^2}} \, dx,x,x^2\right )\\ &=\frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{2} \left (b p^2 q\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {q^2+p^2 x^2}} \, dx,x,x^2\right )+\frac {1}{2} \left (b p q^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {q^2+p^2 x^2}} \, dx,x,x^2\right )\\ &=\frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{2} \left (b p^2 q\right ) \operatorname {Subst}\left (\int \frac {1}{1-p^2 x^2} \, dx,x,\frac {x^2}{\sqrt {q^2+p^2 x^4}}\right )+\frac {1}{4} \left (b p q^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {q^2+p^2 x}} \, dx,x,x^4\right )\\ &=\frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {p x^2}{\sqrt {q^2+p^2 x^4}}\right )+\frac {\left (b q^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {q^2}{p^2}+\frac {x^2}{p^2}} \, dx,x,\sqrt {q^2+p^2 x^4}\right )}{2 p}\\ &=\frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {p x^2}{\sqrt {q^2+p^2 x^4}}\right )-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {\sqrt {q^2+p^2 x^4}}{q}\right )\\ \end {align*}

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Mathematica [C]  time = 0.28, size = 226, normalized size = 2.66 \begin {gather*} \frac {6 a p^2 x^4 \sqrt {p^2 x^4+q^2} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {p^2 x^4}{q^2}\right )+2 a q^2 \sqrt {p^2 x^4+q^2} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {p^2 x^4}{q^2}\right )+3 b p x^3 \sqrt {\frac {p^2 x^4}{q^2}+1} \left (\sqrt {p^2 x^4+q^2}-q \tanh ^{-1}\left (\frac {\sqrt {p^2 x^4+q^2}}{q}\right )\right )+3 b x \sqrt {p^2 x^4+q^2} \left (q \sqrt {\frac {p^2 x^4}{q^2}+1}-p x^2 \sinh ^{-1}\left (\frac {p x^2}{q}\right )\right )}{6 x^3 \sqrt {\frac {p^2 x^4}{q^2}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-q + p*x^2)*(a*q + b*x + a*p*x^2)*Sqrt[q^2 + p^2*x^4])/x^4,x]

[Out]

(3*b*x*Sqrt[q^2 + p^2*x^4]*(q*Sqrt[1 + (p^2*x^4)/q^2] - p*x^2*ArcSinh[(p*x^2)/q]) + 3*b*p*x^3*Sqrt[1 + (p^2*x^
4)/q^2]*(Sqrt[q^2 + p^2*x^4] - q*ArcTanh[Sqrt[q^2 + p^2*x^4]/q]) + 2*a*q^2*Sqrt[q^2 + p^2*x^4]*Hypergeometric2
F1[-3/4, -1/2, 1/4, -((p^2*x^4)/q^2)] + 6*a*p^2*x^4*Sqrt[q^2 + p^2*x^4]*Hypergeometric2F1[-1/2, 1/4, 5/4, -((p
^2*x^4)/q^2)])/(6*x^3*Sqrt[1 + (p^2*x^4)/q^2])

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IntegrateAlgebraic [A]  time = 1.16, size = 85, normalized size = 1.00 \begin {gather*} \frac {\sqrt {q^2+p^2 x^4} \left (2 a q^2+3 b q x+3 b p x^3+2 a p^2 x^4\right )}{6 x^3}+b p q \log (x)-b p q \log \left (q+p x^2+\sqrt {q^2+p^2 x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-q + p*x^2)*(a*q + b*x + a*p*x^2)*Sqrt[q^2 + p^2*x^4])/x^4,x]

[Out]

(Sqrt[q^2 + p^2*x^4]*(2*a*q^2 + 3*b*q*x + 3*b*p*x^3 + 2*a*p^2*x^4))/(6*x^3) + b*p*q*Log[x] - b*p*q*Log[q + p*x
^2 + Sqrt[q^2 + p^2*x^4]]

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fricas [A]  time = 1.49, size = 83, normalized size = 0.98 \begin {gather*} \frac {6 \, b p q x^{3} \log \left (\frac {p x^{2} + q - \sqrt {p^{2} x^{4} + q^{2}}}{x}\right ) + {\left (2 \, a p^{2} x^{4} + 3 \, b p x^{3} + 2 \, a q^{2} + 3 \, b q x\right )} \sqrt {p^{2} x^{4} + q^{2}}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x^2-q)*(a*p*x^2+a*q+b*x)*(p^2*x^4+q^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/6*(6*b*p*q*x^3*log((p*x^2 + q - sqrt(p^2*x^4 + q^2))/x) + (2*a*p^2*x^4 + 3*b*p*x^3 + 2*a*q^2 + 3*b*q*x)*sqrt
(p^2*x^4 + q^2))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {p^{2} x^{4} + q^{2}} {\left (a p x^{2} + a q + b x\right )} {\left (p x^{2} - q\right )}}{x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x^2-q)*(a*p*x^2+a*q+b*x)*(p^2*x^4+q^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(sqrt(p^2*x^4 + q^2)*(a*p*x^2 + a*q + b*x)*(p*x^2 - q)/x^4, x)

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maple [A]  time = 0.25, size = 140, normalized size = 1.65

method	result	size
elliptic	\(\frac {p b \sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {b \,p^{2} q \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}+\frac {b q \sqrt {p^{2} x^{4}+q^{2}}}{2 x^{2}}-\frac {p b \,q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}+\frac {a \left (p^{2} x^{4}+q^{2}\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(140\)
risch	\(\frac {\sqrt {p^{2} x^{4}+q^{2}}\, q \left (2 a q +3 b x \right )}{6 x^{3}}+\frac {a \,p^{2} x \sqrt {p^{2} x^{4}+q^{2}}}{3}+\frac {p b \sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {b \,p^{2} q \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}-\frac {p b \,q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}\)	\(149\)
default	\(a \,p^{2} \left (\frac {x \sqrt {p^{2} x^{4}+q^{2}}}{3}+\frac {2 q^{2} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \EllipticF \left (x \sqrt {\frac {i p}{q}}, i\right )}{3 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )-a \,q^{2} \left (-\frac {\sqrt {p^{2} x^{4}+q^{2}}}{3 x^{3}}+\frac {2 p^{2} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \EllipticF \left (x \sqrt {\frac {i p}{q}}, i\right )}{3 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )+p b \left (\frac {\sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}\right )-q b \left (-\frac {\left (p^{2} x^{4}+q^{2}\right )^{\frac {3}{2}}}{2 q^{2} x^{2}}+\frac {p^{2} x^{2} \sqrt {p^{2} x^{4}+q^{2}}}{2 q^{2}}+\frac {p^{2} \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}\right )\)	\(334\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((p*x^2-q)*(a*p*x^2+a*q+b*x)*(p^2*x^4+q^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/2*p*b*(p^2*x^4+q^2)^(1/2)-1/2*b*p^2*q*ln(p^2*x^2/(p^2)^(1/2)+(p^2*x^4+q^2)^(1/2))/(p^2)^(1/2)+1/2*b*q/x^2*(p
^2*x^4+q^2)^(1/2)-1/2*p*b*q^2/(q^2)^(1/2)*ln((2*q^2+2*(q^2)^(1/2)*(p^2*x^4+q^2)^(1/2))/x^2)+1/3*a*(p^2*x^4+q^2
)^(3/2)/x^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {p^{2} x^{4} + q^{2}} {\left (a p x^{2} + a q + b x\right )} {\left (p x^{2} - q\right )}}{x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x^2-q)*(a*p*x^2+a*q+b*x)*(p^2*x^4+q^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(p^2*x^4 + q^2)*(a*p*x^2 + a*q + b*x)*(p*x^2 - q)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\sqrt {p^2\,x^4+q^2}\,\left (q-p\,x^2\right )\,\left (a\,p\,x^2+b\,x+a\,q\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((p^2*x^4 + q^2)^(1/2)*(q - p*x^2)*(a*q + b*x + a*p*x^2))/x^4,x)

[Out]

-int(((p^2*x^4 + q^2)^(1/2)*(q - p*x^2)*(a*q + b*x + a*p*x^2))/x^4, x)

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sympy [C]  time = 5.45, size = 223, normalized size = 2.62 \begin {gather*} \frac {a p^{2} q x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {p^{2} x^{4} e^{i \pi }}{q^{2}}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} - \frac {a q^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {p^{2} x^{4} e^{i \pi }}{q^{2}}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {b p^{2} x^{2}}{2 \sqrt {\frac {p^{2} x^{4}}{q^{2}} + 1}} + \frac {b p^{2} x^{2}}{2 \sqrt {1 + \frac {q^{2}}{p^{2} x^{4}}}} - \frac {b p q \operatorname {asinh}{\left (\frac {q}{p x^{2}} \right )}}{2} - \frac {b p q \operatorname {asinh}{\left (\frac {p x^{2}}{q} \right )}}{2} + \frac {b q^{2}}{2 x^{2} \sqrt {\frac {p^{2} x^{4}}{q^{2}} + 1}} + \frac {b q^{2}}{2 x^{2} \sqrt {1 + \frac {q^{2}}{p^{2} x^{4}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x**2-q)*(a*p*x**2+a*q+b*x)*(p**2*x**4+q**2)**(1/2)/x**4,x)

[Out]

a*p**2*q*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), p**2*x**4*exp_polar(I*pi)/q**2)/(4*gamma(5/4)) - a*q**3*gamma
(-3/4)*hyper((-3/4, -1/2), (1/4,), p**2*x**4*exp_polar(I*pi)/q**2)/(4*x**3*gamma(1/4)) + b*p**2*x**2/(2*sqrt(p
**2*x**4/q**2 + 1)) + b*p**2*x**2/(2*sqrt(1 + q**2/(p**2*x**4))) - b*p*q*asinh(q/(p*x**2))/2 - b*p*q*asinh(p*x
**2/q)/2 + b*q**2/(2*x**2*sqrt(p**2*x**4/q**2 + 1)) + b*q**2/(2*x**2*sqrt(1 + q**2/(p**2*x**4)))

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