3.9.26 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{(e x)^{5/2}} \, dx\) [826]

Optimal. Leaf size=234 \[ -\frac {2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{21 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac {2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{c} d^{5/4} e^{5/2} \sqrt {c+d x^2}} \]

[Out]

-2/3*a^2*(d*x^2+c)^(3/2)/c/e/(e*x)^(3/2)+2/7*b^2*(d*x^2+c)^(3/2)*(e*x)^(1/2)/d/e^3-2/21*(b^2*c^2-7*a*d*(a*d+2*
b*c))*(e*x)^(1/2)*(d*x^2+c)^(1/2)/c/d/e^3-2/21*(b^2*c^2-7*a*d*(a*d+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^(1/4)/d
^(5/4)/e^(5/2)/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 234, 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 = {473, 470, 285, 335, 226} \begin {gather*} -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}-\frac {2 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (b^2 c^2-7 a d (a d+2 b c)\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{c} d^{5/4} e^{5/2} \sqrt {c+d x^2}}-\frac {2 \sqrt {e x} \sqrt {c+d x^2} \left (b^2 c^2-7 a d (a d+2 b c)\right )}{21 c d e^3}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{3/2}}{7 d e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/(e*x)^(5/2),x]

[Out]

(-2*(b^2*c^2 - 7*a*d*(2*b*c + a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(21*c*d*e^3) - (2*a^2*(c + d*x^2)^(3/2))/(3*c*e
*(e*x)^(3/2)) + (2*b^2*Sqrt[e*x]*(c + d*x^2)^(3/2))/(7*d*e^3) - (2*(b^2*c^2 - 7*a*d*(2*b*c + 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])/(21*c^(1/4)*d^(5/4)*e^(5/2)*Sqrt[c + d*x^2])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 285

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

Rule 335

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 470

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

Rule 473

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 \sqrt {c+d x^2}}{(e x)^{5/2}} \, dx &=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac {2 \int \frac {\left (\frac {3}{2} a (2 b c+a d)+\frac {3}{2} b^2 c x^2\right ) \sqrt {c+d x^2}}{\sqrt {e x}} \, dx}{3 c e^2}\\ &=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac {\left (b^2 c^2-7 a d (2 b c+a d)\right ) \int \frac {\sqrt {c+d x^2}}{\sqrt {e x}} \, dx}{7 c d e^2}\\ &=-\frac {2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{21 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac {\left (2 \left (b^2 c^2-7 a d (2 b c+a d)\right )\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{21 d e^2}\\ &=-\frac {2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{21 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac {\left (4 \left (b^2 c^2-7 a d (2 b c+a d)\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{21 d e^3}\\ &=-\frac {2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{21 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac {2 \left (b^2 c^2-7 a d (2 b c+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 )}{21 \sqrt [4]{c} d^{5/4} e^{5/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.14, size = 171, normalized size = 0.73 \begin {gather*} \frac {x^{5/2} \left (\frac {2 \left (c+d x^2\right ) \left (-7 a^2 d+14 a b d x^2+b^2 x^2 \left (2 c+3 d x^2\right )\right )}{d x^{3/2}}+\frac {4 i \left (-b^2 c^2+14 a b c d+7 a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} d}\right )}{21 (e x)^{5/2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/(e*x)^(5/2),x]

[Out]

(x^(5/2)*((2*(c + d*x^2)*(-7*a^2*d + 14*a*b*d*x^2 + b^2*x^2*(2*c + 3*d*x^2)))/(d*x^(3/2)) + ((4*I)*(-(b^2*c^2)
+ 14*a*b*c*d + 7*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/
(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d)))/(21*(e*x)^(5/2)*Sqrt[c + d*x^2])

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Maple [A]
time = 0.12, size = 383, normalized size = 1.64

method	result	size
risch	\(-\frac {2 \sqrt {d \,x^{2}+c}\, \left (-3 b^{2} d \,x^{4}-14 a b d \,x^{2}-2 b^{2} c \,x^{2}+7 a^{2} d \right )}{21 d x \,e^{2} \sqrt {e x}}+\frac {2 \left (7 a^{2} d^{2}+14 a b c d -b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{21 d^{2} \sqrt {d e \,x^{3}+c e x}\, e^{2} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(222\)
elliptic	\(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{3 e^{3} x^{2}}+\frac {2 b^{2} x^{2} \sqrt {d e \,x^{3}+c e x}}{7 e^{3}}+\frac {2 \left (\frac {b \left (2 a d +b c \right )}{e^{2}}-\frac {5 b^{2} c}{7 e^{2}}\right ) \sqrt {d e \,x^{3}+c e x}}{3 d e}+\frac {\left (\frac {a \left (a d +2 b c \right )}{e^{2}}-\frac {d \,a^{2}}{3 e^{2}}-\frac {\left (\frac {b \left (2 a d +b c \right )}{e^{2}}-\frac {5 b^{2} c}{7 e^{2}}\right ) c}{3 d}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(282\)
default	\(\frac {\frac {2 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} d^{2} x}{3}+\frac {4 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b c d x}{3}-\frac {2 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{2} x}{21}+\frac {2 b^{2} x^{6} d^{3}}{7}+\frac {4 a b \,d^{3} x^{4}}{3}+\frac {10 b^{2} c \,d^{2} x^{4}}{21}-\frac {2 a^{2} d^{3} x^{2}}{3}+\frac {4 a b c \,d^{2} x^{2}}{3}+\frac {4 b^{2} c^{2} d \,x^{2}}{21}-\frac {2 a^{2} c \,d^{2}}{3}}{\sqrt {d \,x^{2}+c}\, x \,e^{2} \sqrt {e x}\, d^{2}}\)	\(383\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/21/(d*x^2+c)^(1/2)/x*(7*(-c*d)^(1/2)*((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)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2
*d^2*x+14*(-c*d)^(1/2)*((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
)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c*d*x-(-c*d)^(1
/2)*((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)*(-x/(-c*d)^(1/2)*
d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^2*x+3*b^2*x^6*d^3+14*a*b*d^3*x^4
+5*b^2*c*d^2*x^4-7*a^2*d^3*x^2+14*a*b*c*d^2*x^2+2*b^2*c^2*d*x^2-7*a^2*c*d^2)/e^2/(e*x)^(1/2)/d^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/x^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.36, size = 103, normalized size = 0.44 \begin {gather*} -\frac {2 \, {\left (2 \, {\left (b^{2} c^{2} - 14 \, a b c d - 7 \, a^{2} d^{2}\right )} \sqrt {d} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right ) - {\left (3 \, b^{2} d^{2} x^{4} - 7 \, a^{2} d^{2} + 2 \, {\left (b^{2} c d + 7 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {x}\right )} e^{\left (-\frac {5}{2}\right )}}{21 \, d^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

-2/21*(2*(b^2*c^2 - 14*a*b*c*d - 7*a^2*d^2)*sqrt(d)*x^2*weierstrassPInverse(-4*c/d, 0, x) - (3*b^2*d^2*x^4 - 7
*a^2*d^2 + 2*(b^2*c*d + 7*a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(x))*e^(-5/2)/(d^2*x^2)

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Sympy [C] Result contains complex when optimal does not.
time = 5.91, size = 153, normalized size = 0.65 \begin {gather*} \frac {a^{2} \sqrt {c} \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 {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {a b \sqrt {c} \sqrt {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 {d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {b^{2} \sqrt {c} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/(e*x)**(5/2),x)

[Out]

a**2*sqrt(c)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*x**(3/2)*gamma(1/4)
) + a*b*sqrt(c)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), d*x**2*exp_polar(I*pi)/c)/(e**(5/2)*gamma(5/4))
+ b**2*sqrt(c)*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*gamma(9/4)
)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)*e^(-5/2)/x^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{{\left (e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/(e*x)^(5/2),x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/(e*x)^(5/2), x)

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