3.1126 \(\int \frac{(e x)^{7/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=221 \[ -\frac{e^3 \sqrt{e x} (4 b c-9 a d)}{2 b^3 \sqrt [4]{a+b x^2}}+\frac{e^{7/2} (4 b c-9 a d) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}+\frac{e^{7/2} (4 b c-9 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}-\frac{e (e x)^{5/2} (4 b c-9 a d)}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

[Out]

(2*(b*c - a*d)*(e*x)^(9/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - ((4*b*c - 9*a*d)*e^3*Sqrt[e*x])/(2*b^3*(a + b*x^2)^(
1/4)) - ((4*b*c - 9*a*d)*e*(e*x)^(5/2))/(10*a*b^2*(a + b*x^2)^(1/4)) + ((4*b*c - 9*a*d)*e^(7/2)*ArcTan[(b^(1/4
)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(13/4)) + ((4*b*c - 9*a*d)*e^(7/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])
/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(13/4))

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Rubi [A]  time = 0.13475, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {457, 285, 288, 329, 240, 212, 208, 205} \[ -\frac{e^3 \sqrt{e x} (4 b c-9 a d)}{2 b^3 \sqrt [4]{a+b x^2}}+\frac{e^{7/2} (4 b c-9 a d) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}+\frac{e^{7/2} (4 b c-9 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}-\frac{e (e x)^{5/2} (4 b c-9 a d)}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(9/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - ((4*b*c - 9*a*d)*e^3*Sqrt[e*x])/(2*b^3*(a + b*x^2)^(
1/4)) - ((4*b*c - 9*a*d)*e*(e*x)^(5/2))/(10*a*b^2*(a + b*x^2)^(1/4)) + ((4*b*c - 9*a*d)*e^(7/2)*ArcTan[(b^(1/4
)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(13/4)) + ((4*b*c - 9*a*d)*e^(7/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])
/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(13/4))

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 285

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(2*c*(c*x)^(m - 1))/(b*(2*m - 3)*(a + b*x
^2)^(1/4)), x] - Dist[(2*a*c^2*(m - 1))/(b*(2*m - 3)), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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 240

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac{\left (2 \left (-2 b c+\frac{9 a d}{2}\right )\right ) \int \frac{(e x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a b}\\ &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{(4 b c-9 a d) e (e x)^{5/2}}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-9 a d) e^2\right ) \int \frac{(e x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{4 b^2}\\ &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{(4 b c-9 a d) e^3 \sqrt{e x}}{2 b^3 \sqrt [4]{a+b x^2}}-\frac{(4 b c-9 a d) e (e x)^{5/2}}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-9 a d) e^4\right ) \int \frac{1}{\sqrt{e x} \sqrt [4]{a+b x^2}} \, dx}{4 b^3}\\ &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{(4 b c-9 a d) e^3 \sqrt{e x}}{2 b^3 \sqrt [4]{a+b x^2}}-\frac{(4 b c-9 a d) e (e x)^{5/2}}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-9 a d) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{2 b^3}\\ &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{(4 b c-9 a d) e^3 \sqrt{e x}}{2 b^3 \sqrt [4]{a+b x^2}}-\frac{(4 b c-9 a d) e (e x)^{5/2}}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-9 a d) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{b x^4}{e^2}} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{2 b^3}\\ &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{(4 b c-9 a d) e^3 \sqrt{e x}}{2 b^3 \sqrt [4]{a+b x^2}}-\frac{(4 b c-9 a d) e (e x)^{5/2}}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-9 a d) e^4\right ) \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{b} x^2} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{4 b^3}+\frac{\left ((4 b c-9 a d) e^4\right ) \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{b} x^2} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{4 b^3}\\ &=\frac{2 (b c-a d) (e x)^{9/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{(4 b c-9 a d) e^3 \sqrt{e x}}{2 b^3 \sqrt [4]{a+b x^2}}-\frac{(4 b c-9 a d) e (e x)^{5/2}}{10 a b^2 \sqrt [4]{a+b x^2}}+\frac{(4 b c-9 a d) e^{7/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}+\frac{(4 b c-9 a d) e^{7/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}\\ \end{align*}

Mathematica [C]  time = 0.115709, size = 91, normalized size = 0.41 \[ \frac{e^3 x^4 \sqrt{e x} \left (9 a^2 d+\left (a+b x^2\right ) \sqrt [4]{\frac{b x^2}{a}+1} (4 b c-9 a d) \, _2F_1\left (\frac{9}{4},\frac{9}{4};\frac{13}{4};-\frac{b x^2}{a}\right )\right )}{18 a^2 b \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(e^3*x^4*Sqrt[e*x]*(9*a^2*d + (4*b*c - 9*a*d)*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[9/4, 9/4, 13
/4, -((b*x^2)/a)]))/(18*a^2*b*(a + b*x^2)^(5/4))

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c) \left ( ex \right ) ^{{\frac{7}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{9}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.46799, size = 2306, normalized size = 10.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

1/40*(4*(5*b^2*d*e^3*x^4 - 6*(4*b^2*c - 9*a*b*d)*e^3*x^2 - 5*(4*a*b*c - 9*a^2*d)*e^3)*(b*x^2 + a)^(3/4)*sqrt(e
*x) + 20*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3
*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4)*arctan(((4*b^11*c - 9*a*b^10*d)*(b*x^2 + a)^(3/4)*((256*b^4*c^4 - 23
04*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(3/4)*sqrt(e*x)*e^3 + (b^
11*x^2 + a*b^10)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e
^14/b^13)^(3/4)*sqrt(((16*b^2*c^2 - 72*a*b*c*d + 81*a^2*d^2)*sqrt(b*x^2 + a)*e^7*x + (b^7*x^2 + a*b^6)*sqrt((2
56*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13))/(b*x^2 +
a)))/((256*b^5*c^4 - 2304*a*b^4*c^3*d + 7776*a^2*b^3*c^2*d^2 - 11664*a^3*b^2*c*d^3 + 6561*a^4*b*d^4)*e^14*x^2
+ (256*a*b^4*c^4 - 2304*a^2*b^3*c^3*d + 7776*a^3*b^2*c^2*d^2 - 11664*a^4*b*c*d^3 + 6561*a^5*d^4)*e^14)) + 5*(b
^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 +
6561*a^4*d^4)*e^14/b^13)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 9*a*d)*sqrt(e*x)*e^3 + (b^4*x^2 + a*b^3)*((256
*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4))/(b*x^
2 + a)) - 5*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*
a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 9*a*d)*sqrt(e*x)*e^3 - (b^4*x^2
+ a*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13
)^(1/4))/(b*x^2 + a)))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

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