3.9.80 \(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=149 \[ \frac {d e^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {2 (e x)^{5/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

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Rubi [A]  time = 0.09, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {452, 288, 329, 240, 212, 208, 205} \begin {gather*} \frac {d e^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {2 (e x)^{5/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

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

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]

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 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 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 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 452

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*(m + 1)), x] + Dist[d/b, Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /;
FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 77, normalized size = 0.52 \begin {gather*} \frac {2 x (e x)^{3/2} \left (5 d x^2 \left (a+b x^2\right ) \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {9}{4},\frac {9}{4};\frac {13}{4};-\frac {b x^2}{a}\right )+9 a c\right )}{45 a^2 \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(2*x*(e*x)^(3/2)*(9*a*c + 5*d*x^2*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[9/4, 9/4, 13/4, -((b*x^2
)/a)]))/(45*a^2*(a + b*x^2)^(5/4))

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IntegrateAlgebraic [A]  time = 28.61, size = 206, normalized size = 1.38 \begin {gather*} \frac {e^{3/2} \left (a+b x^2\right )^{3/4} \left (-\frac {2 \left (5 a^2 d e^{7/2} \sqrt {e x}+6 a b d e^{3/2} (e x)^{5/2}-b^2 c e^{3/2} (e x)^{5/2}\right )}{5 a b^2 \left (a e^2+b e^2 x^2\right )^{5/4}}+\frac {d e^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a e^2+b e^2 x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a e^2+b e^2 x^2}}\right )}{b^{9/4}}\right )}{\left (a e^2+b e^2 x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(e^(3/2)*(a + b*x^2)^(3/4)*((-2*(5*a^2*d*e^(7/2)*Sqrt[e*x] - b^2*c*e^(3/2)*(e*x)^(5/2) + 6*a*b*d*e^(3/2)*(e*x)
^(5/2)))/(5*a*b^2*(a*e^2 + b*e^2*x^2)^(5/4)) + (d*e^(3/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(a*e^2 + b*e^2*x^2)^(1/4)
])/b^(9/4) + (d*e^(3/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(a*e^2 + b*e^2*x^2)^(1/4)])/b^(9/4)))/(a*e^2 + b*e^2*x^2)^
(3/4)

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fricas [B]  time = 1.53, size = 448, normalized size = 3.01 \begin {gather*} -\frac {4 \, {\left (5 \, a^{2} d e - {\left (b^{2} c - 6 \, a b d\right )} e x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} + 20 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} b^{7} d e \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b^{8} x^{2} + a b^{7}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {3}{4}} \sqrt {\frac {\sqrt {b x^{2} + a} d^{2} e^{3} x + {\left (b^{5} x^{2} + a b^{4}\right )} \sqrt {\frac {d^{4} e^{6}}{b^{9}}}}{b x^{2} + a}}}{b d^{4} e^{6} x^{2} + a d^{4} e^{6}}\right ) - 5 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d e + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + 5 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d e - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right )}{10 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

-1/10*(4*(5*a^2*d*e - (b^2*c - 6*a*b*d)*e*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x) + 20*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a
^3*b^2)*(d^4*e^6/b^9)^(1/4)*arctan(-((b*x^2 + a)^(3/4)*sqrt(e*x)*b^7*d*e*(d^4*e^6/b^9)^(3/4) - (b^8*x^2 + a*b^
7)*(d^4*e^6/b^9)^(3/4)*sqrt((sqrt(b*x^2 + a)*d^2*e^3*x + (b^5*x^2 + a*b^4)*sqrt(d^4*e^6/b^9))/(b*x^2 + a)))/(b
*d^4*e^6*x^2 + a*d^4*e^6)) - 5*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(d^4*e^6/b^9)^(1/4)*log(((b*x^2 + a)^(3/4
)*sqrt(e*x)*d*e + (b^3*x^2 + a*b^2)*(d^4*e^6/b^9)^(1/4))/(b*x^2 + a)) + 5*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2
)*(d^4*e^6/b^9)^(1/4)*log(((b*x^2 + a)^(3/4)*sqrt(e*x)*d*e - (b^3*x^2 + a*b^2)*(d^4*e^6/b^9)^(1/4))/(b*x^2 + a
)))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(9/4), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

[Out]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(9/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^{3/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x)

[Out]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(d*x**2+c)/(b*x**2+a)**(9/4),x)

[Out]

Timed out

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