3.1104 \(\int \frac{c+d x^2}{(e x)^{13/2} (a+b x^2)^{3/4}} \, dx\)

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

[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))

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Rubi [A]  time = 0.140598, antiderivative size = 182, normalized size of antiderivative = 1., 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 = {453, 325, 329, 237, 335, 275, 231} \[ \frac{8 b^{5/2} (e x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} (10 b c-11 a d) F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{77 a^{7/2} e^8 \left (a+b x^2\right )^{3/4}}-\frac{4 b \sqrt [4]{a+b x^2} (10 b c-11 a d)}{77 a^3 e^5 (e x)^{3/2}}+\frac{2 \sqrt [4]{a+b x^2} (10 b c-11 a d)}{77 a^2 e^3 (e x)^{7/2}}-\frac{2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}} \]

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 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 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 237

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 335

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 275

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 231

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

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.0684601, size = 88, normalized size = 0.48 \[ -\frac{2 \sqrt{e x} \left (x^2 \left (\frac{b x^2}{a}+1\right )^{3/4} (11 a d-10 b c) \, _2F_1\left (-\frac{7}{4},\frac{3}{4};-\frac{3}{4};-\frac{b x^2}{a}\right )+7 c \left (a+b x^2\right )\right )}{77 a e^7 x^6 \left (a+b x^2\right )^{3/4}} \]

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 = 0.031, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c) \left ( ex \right ) ^{-{\frac{13}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \left (e x\right )^{\frac{13}{2}}}\,{d x} \end{align*}

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]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{1}{4}}{\left (d x^{2} + c\right )} \sqrt{e x}}{b e^{7} x^{9} + a e^{7} x^{7}}, x\right ) \end{align*}

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(e*x)/(b*e^7*x^9 + a*e^7*x^7), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \left (e x\right )^{\frac{13}{2}}}\,{d x} \end{align*}

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