3.242 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=204 \[ \frac {4 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 A c+7 b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{b} \sqrt {b x^2+c x^4}}+\frac {4 c \sqrt {b x^2+c x^4} (3 A c+7 b B)}{21 b \sqrt {x}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (3 A c+7 b B)}{21 b x^{9/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}} \]

[Out]

-2/21*(3*A*c+7*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^(9/2)-2/7*A*(c*x^4+b*x^2)^(5/2)/b/x^(17/2)+4/21*c*(3*A*c+7*B*b)*(c
*x^4+b*x^2)^(1/2)/b/x^(1/2)+4/21*c^(3/4)*(3*A*c+7*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(
2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^
(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/b^(1/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2038, 2020, 2021, 2032, 329, 220} \[ \frac {4 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 A c+7 b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{b} \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (3 A c+7 b B)}{21 b x^{9/2}}+\frac {4 c \sqrt {b x^2+c x^4} (3 A c+7 b B)}{21 b \sqrt {x}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(15/2),x]

[Out]

(4*c*(7*b*B + 3*A*c)*Sqrt[b*x^2 + c*x^4])/(21*b*Sqrt[x]) - (2*(7*b*B + 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(21*b*x^(
9/2)) - (2*A*(b*x^2 + c*x^4)^(5/2))/(7*b*x^(17/2)) + (4*c^(3/4)*(7*b*B + 3*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(
b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(21*b^(1/4)*Sqrt[b*x^
2 + c*x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx &=-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}}-\frac {\left (2 \left (-\frac {7 b B}{2}-\frac {3 A c}{2}\right )\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx}{7 b}\\ &=-\frac {2 (7 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b x^{9/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}}+\frac {(2 c (7 b B+3 A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx}{7 b}\\ &=\frac {4 c (7 b B+3 A c) \sqrt {b x^2+c x^4}}{21 b \sqrt {x}}-\frac {2 (7 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b x^{9/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}}+\frac {1}{21} (4 c (7 b B+3 A c)) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {4 c (7 b B+3 A c) \sqrt {b x^2+c x^4}}{21 b \sqrt {x}}-\frac {2 (7 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b x^{9/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}}+\frac {\left (4 c (7 b B+3 A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{21 \sqrt {b x^2+c x^4}}\\ &=\frac {4 c (7 b B+3 A c) \sqrt {b x^2+c x^4}}{21 b \sqrt {x}}-\frac {2 (7 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b x^{9/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}}+\frac {\left (8 c (7 b B+3 A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{21 \sqrt {b x^2+c x^4}}\\ &=\frac {4 c (7 b B+3 A c) \sqrt {b x^2+c x^4}}{21 b \sqrt {x}}-\frac {2 (7 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b x^{9/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}}+\frac {4 c^{3/4} (7 b B+3 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{b} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 101, normalized size = 0.50 \[ -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (b x^2 (3 A c+7 b B) \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {c x^2}{b}\right )+3 A \left (b+c x^2\right )^2 \sqrt {\frac {c x^2}{b}+1}\right )}{21 b x^{9/2} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(15/2),x]

[Out]

(-2*Sqrt[x^2*(b + c*x^2)]*(3*A*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] + b*(7*b*B + 3*A*c)*x^2*Hypergeometric2F1[-3/
2, -3/4, 1/4, -((c*x^2)/b)]))/(21*b*x^(9/2)*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c x^{4} + {\left (B b + A c\right )} x^{2} + A b\right )} \sqrt {c x^{4} + b x^{2}}}{x^{\frac {11}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="fricas")

[Out]

integral((B*c*x^4 + (B*b + A*c)*x^2 + A*b)*sqrt(c*x^4 + b*x^2)/x^(11/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2), x)

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maple [A]  time = 0.08, size = 254, normalized size = 1.25 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (7 B \,c^{2} x^{6}-9 A \,c^{2} x^{4}+6 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A c \,x^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+14 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B b \,x^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-12 A b c \,x^{2}-7 B \,b^{2} x^{2}-3 A \,b^{2}\right )}{21 \left (c \,x^{2}+b \right )^{2} x^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(15/2),x)

[Out]

2/21*(c*x^4+b*x^2)^(3/2)/x^(13/2)/(c*x^2+b)^2*(6*A*(-b*c)^(1/2)*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2
)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1
/2))^(1/2),1/2*2^(1/2))*x^3*c+14*B*(-b*c)^(1/2)*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^
(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2
^(1/2))*x^3*b+7*B*c^2*x^6-9*A*c^2*x^4-12*A*b*c*x^2-7*B*b^2*x^2-3*A*b^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{15/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(15/2),x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(15/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**(15/2),x)

[Out]

Timed out

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