3.848 \(\int \frac {(a+b x^2)^2}{(e x)^{13/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=242 \[ -\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}-\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 c^{13/4} e^{13/2} \sqrt {c+d x^2}}-\frac {2 \sqrt {c+d x^2} \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right )}{231 c^3 e^5 (e x)^{3/2}}-\frac {2 a \sqrt {c+d x^2} (22 b c-9 a d)}{77 c^2 e^3 (e x)^{7/2}} \]

[Out]

-2/11*a^2*(d*x^2+c)^(1/2)/c/e/(e*x)^(11/2)-2/77*a*(-9*a*d+22*b*c)*(d*x^2+c)^(1/2)/c^2/e^3/(e*x)^(7/2)-2/231*(7
7*b^2*c^2-5*a*d*(-9*a*d+22*b*c))*(d*x^2+c)^(1/2)/c^3/e^5/(e*x)^(3/2)-1/231*d^(3/4)*(77*b^2*c^2-5*a*d*(-9*a*d+2
2*b*c))*(cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/
e^(1/2)))*EllipticF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^
2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/c^(13/4)/e^(13/2)/(d*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {462, 453, 325, 329, 220} \[ -\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}-\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 c^{13/4} e^{13/2} \sqrt {c+d x^2}}-\frac {2 \sqrt {c+d x^2} \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right )}{231 c^3 e^5 (e x)^{3/2}}-\frac {2 a \sqrt {c+d x^2} (22 b c-9 a d)}{77 c^2 e^3 (e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(13/2)*Sqrt[c + d*x^2]),x]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(11*c*e*(e*x)^(11/2)) - (2*a*(22*b*c - 9*a*d)*Sqrt[c + d*x^2])/(77*c^2*e^3*(e*x)^(7/2
)) - (2*(77*b^2*c^2 - 5*a*d*(22*b*c - 9*a*d))*Sqrt[c + d*x^2])/(231*c^3*e^5*(e*x)^(3/2)) - (d^(3/4)*(77*b^2*c^
2 - 5*a*d*(22*b*c - 9*a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan
[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(231*c^(13/4)*e^(13/2)*Sqrt[c + d*x^2])

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 453

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] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{(e x)^{13/2} \sqrt {c+d x^2}} \, dx &=-\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}+\frac {2 \int \frac {\frac {1}{2} a (22 b c-9 a d)+\frac {11}{2} b^2 c x^2}{(e x)^{9/2} \sqrt {c+d x^2}} \, dx}{11 c e^2}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}-\frac {2 a (22 b c-9 a d) \sqrt {c+d x^2}}{77 c^2 e^3 (e x)^{7/2}}+\frac {\left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) \int \frac {1}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx}{77 c^2 e^4}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}-\frac {2 a (22 b c-9 a d) \sqrt {c+d x^2}}{77 c^2 e^3 (e x)^{7/2}}-\frac {2 \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) \sqrt {c+d x^2}}{231 c^3 e^5 (e x)^{3/2}}-\frac {\left (d \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right )\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{231 c^3 e^6}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}-\frac {2 a (22 b c-9 a d) \sqrt {c+d x^2}}{77 c^2 e^3 (e x)^{7/2}}-\frac {2 \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) \sqrt {c+d x^2}}{231 c^3 e^5 (e x)^{3/2}}-\frac {\left (2 d \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{231 c^3 e^7}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{11 c e (e x)^{11/2}}-\frac {2 a (22 b c-9 a d) \sqrt {c+d x^2}}{77 c^2 e^3 (e x)^{7/2}}-\frac {2 \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) \sqrt {c+d x^2}}{231 c^3 e^5 (e x)^{3/2}}-\frac {d^{3/4} \left (77 b^2 c^2-5 a d (22 b c-9 a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 c^{13/4} e^{13/2} \sqrt {c+d x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.26, size = 196, normalized size = 0.81 \[ \frac {x^{13/2} \left (-\frac {2 \left (c+d x^2\right ) \left (3 a^2 \left (7 c^2-9 c d x^2+15 d^2 x^4\right )+22 a b c x^2 \left (3 c-5 d x^2\right )+77 b^2 c^2 x^4\right )}{c^3 x^{11/2}}-\frac {2 i d x \sqrt {\frac {c}{d x^2}+1} \left (45 a^2 d^2-110 a b c d+77 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{c^3 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}\right )}{231 (e x)^{13/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(13/2)*Sqrt[c + d*x^2]),x]

[Out]

(x^(13/2)*((-2*(c + d*x^2)*(77*b^2*c^2*x^4 + 22*a*b*c*x^2*(3*c - 5*d*x^2) + 3*a^2*(7*c^2 - 9*c*d*x^2 + 15*d^2*
x^4)))/(c^3*x^(11/2)) - ((2*I)*d*(77*b^2*c^2 - 110*a*b*c*d + 45*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*Arc
Sinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(c^3*Sqrt[(I*Sqrt[c])/Sqrt[d]])))/(231*(e*x)^(13/2)*Sqrt[c + d*x
^2])

________________________________________________________________________________________

fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{d e^{7} x^{9} + c e^{7} x^{7}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(13/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(d*e^7*x^9 + c*e^7*x^7), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(13/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(13/2)), x)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 411, normalized size = 1.70 \[ -\frac {90 a^{2} d^{3} x^{6}-220 a b c \,d^{2} x^{6}+154 b^{2} c^{2} d \,x^{6}+45 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a^{2} d^{2} x^{5} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-110 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a b c d \,x^{5} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+77 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, b^{2} c^{2} x^{5} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+36 a^{2} c \,d^{2} x^{4}-88 a b \,c^{2} d \,x^{4}+154 b^{2} c^{3} x^{4}-12 a^{2} c^{2} d \,x^{2}+132 a b \,c^{3} x^{2}+42 a^{2} c^{3}}{231 \sqrt {d \,x^{2}+c}\, \sqrt {e x}\, c^{3} e^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(e*x)^(13/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/231/(d*x^2+c)^(1/2)/x^5*(45*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/
2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1
/2)*x^5*a^2*d^2-110*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(
-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*x^5*a*b
*c*d+77*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1
/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*x^5*b^2*c^2+90*a^2*
d^3*x^6-220*a*b*c*d^2*x^6+154*b^2*c^2*d*x^6+36*a^2*c*d^2*x^4-88*a*b*c^2*d*x^4+154*b^2*c^3*x^4-12*a^2*c^2*d*x^2
+132*a*b*c^3*x^2+42*a^2*c^3)/c^3/e^6/(e*x)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(13/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(13/2)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{13/2}\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/((e*x)^(13/2)*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^2/((e*x)^(13/2)*(c + d*x^2)^(1/2)), x)

________________________________________________________________________________________

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)**2/(e*x)**(13/2)/(d*x**2+c)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________