3.12.4 \(\int \frac {c+d x^2}{(e x)^{13/2} (a+b x^2)^{3/4}} \, dx\) [1104]

Optimal. Leaf size=182 \[ -\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}+\frac {8 b^{5/2} (10 b c-11 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{77 a^{7/2} e^8 \left (a+b x^2\right )^{3/4}} \]

[Out]

-2/11*c*(b*x^2+a)^(1/4)/a/e/(e*x)^(11/2)+2/77*(-11*a*d+10*b*c)*(b*x^2+a)^(1/4)/a^2/e^3/(e*x)^(7/2)-4/77*b*(-11
*a*d+10*b*c)*(b*x^2+a)^(1/4)/a^3/e^5/(e*x)^(3/2)+8/77*b^(5/2)*(-11*a*d+10*b*c)*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*(
cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x*b^(1
/2)/a^(1/2))),2^(1/2))/a^(7/2)/e^8/(b*x^2+a)^(3/4)

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Rubi [A]
time = 0.09, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {464, 331, 335, 243, 342, 281, 237} \begin {gather*} \frac {8 b^{5/2} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (10 b c-11 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{77 a^{7/2} e^8 \left (a+b x^2\right )^{3/4}}-\frac {4 b \sqrt [4]{a+b x^2} (10 b c-11 a d)}{77 a^3 e^5 (e x)^{3/2}}+\frac {2 \sqrt [4]{a+b x^2} (10 b c-11 a d)}{77 a^2 e^3 (e x)^{7/2}}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*c*(a + b*x^2)^(1/4))/(11*a*e*(e*x)^(11/2)) + (2*(10*b*c - 11*a*d)*(a + b*x^2)^(1/4))/(77*a^2*e^3*(e*x)^(7/
2)) - (4*b*(10*b*c - 11*a*d)*(a + b*x^2)^(1/4))/(77*a^3*e^5*(e*x)^(3/2)) + (8*b^(5/2)*(10*b*c - 11*a*d)*(1 + a
/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(77*a^(7/2)*e^8*(a + b*x^2)^(3/4))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 281

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

Rule 331

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 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 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 464

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]

Rubi steps

\begin {align*} \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}-\frac {(10 b c-11 a d) \int \frac {1}{(e x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx}{11 a e^2}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}+\frac {(6 b (10 b c-11 a d)) \int \frac {1}{(e x)^{5/2} \left (a+b x^2\right )^{3/4}} \, dx}{77 a^2 e^4}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}-\frac {\left (4 b^2 (10 b c-11 a d)\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{77 a^3 e^6}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}-\frac {\left (8 b^2 (10 b c-11 a d)\right ) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{77 a^3 e^7}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}-\frac {\left (8 b^2 (10 b c-11 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {e x}\right )}{77 a^3 e^7 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}+\frac {\left (8 b^2 (10 b c-11 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a e^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{77 a^3 e^7 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}+\frac {\left (4 b^2 (10 b c-11 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{e x}\right )}{77 a^3 e^7 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}+\frac {8 b^{5/2} (10 b c-11 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{77 a^{7/2} e^8 \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.06, size = 88, normalized size = 0.48 \begin {gather*} -\frac {2 \sqrt {e x} \left (7 c \left (a+b x^2\right )+(-10 b c+11 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (-\frac {7}{4},\frac {3}{4};-\frac {3}{4};-\frac {b x^2}{a}\right )\right )}{77 a e^7 x^6 \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*Sqrt[e*x]*(7*c*(a + b*x^2) + (-10*b*c + 11*a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[-7/4, 3/4, -3/
4, -((b*x^2)/a)]))/(77*a*e^7*x^6*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {13}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x)

[Out]

int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x)

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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((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

e^(-13/2)*integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*x^(13/2)), x)

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Fricas [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((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)*sqrt(x)*e^(-13/2)/(b*x^9 + a*x^7), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(e*x)**(13/2)/(b*x**2+a)**(3/4),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4063 deep

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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((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*e^(-13/2)/((b*x^2 + a)^(3/4)*x^(13/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {d\,x^2+c}{{\left (e\,x\right )}^{13/2}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)),x)

[Out]

int((c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)), x)

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