3.1105 \(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=171 \[ \frac {e^{3/2} (4 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {e^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac {e \sqrt {e x} (4 b c-5 a d)}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}} \]

[Out]

1/2*d*(e*x)^(5/2)/b/e/(b*x^2+a)^(1/4)+1/4*(-5*a*d+4*b*c)*e^(3/2)*arctan(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^
(1/2))/b^(9/4)+1/4*(-5*a*d+4*b*c)*e^(3/2)*arctanh(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))/b^(9/4)-1/2*(-5
*a*d+4*b*c)*e*(e*x)^(1/2)/b^2/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.11, antiderivative size = 171, 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 = {459, 288, 329, 240, 212, 208, 205} \[ \frac {e^{3/2} (4 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {e^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac {e \sqrt {e x} (4 b c-5 a d)}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

-((4*b*c - 5*a*d)*e*Sqrt[e*x])/(2*b^2*(a + b*x^2)^(1/4)) + (d*(e*x)^(5/2))/(2*b*e*(a + b*x^2)^(1/4)) + ((4*b*c
- 5*a*d)*e^(3/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(9/4)) + ((4*b*c - 5*a*d)*e^(3
/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(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 459

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]

Rubi steps

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

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Mathematica [C]  time = 0.12, size = 77, normalized size = 0.45 \[ \frac {x (e x)^{3/2} \left (\sqrt [4]{\frac {b x^2}{a}+1} (4 b c-5 a d) \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};-\frac {b x^2}{a}\right )+5 a d\right )}{10 a b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(x*(e*x)^(3/2)*(5*a*d + (4*b*c - 5*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((b*x^2)/a)]))
/(10*a*b*(a + b*x^2)^(1/4))

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fricas [B]  time = 0.79, size = 916, normalized size = 5.36 \[ \frac {4 \, {\left (b d e x^{2} - {\left (4 \, b c - 5 \, a d\right )} e\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} + 4 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (4 \, b^{8} c - 5 \, a b^{7} d\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} e \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b^{8} x^{2} + a b^{7}\right )} \sqrt {\frac {{\left (16 \, b^{2} c^{2} - 40 \, a b c d + 25 \, a^{2} d^{2}\right )} \sqrt {b x^{2} + a} e^{3} x + {\left (b^{5} x^{2} + a b^{4}\right )} \sqrt {\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}}}{b x^{2} + a}} \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {3}{4}}}{{\left (256 \, b^{5} c^{4} - 1280 \, a b^{4} c^{3} d + 2400 \, a^{2} b^{3} c^{2} d^{2} - 2000 \, a^{3} b^{2} c d^{3} + 625 \, a^{4} b d^{4}\right )} e^{6} x^{2} + {\left (256 \, a b^{4} c^{4} - 1280 \, a^{2} b^{3} c^{3} d + 2400 \, a^{3} b^{2} c^{2} d^{2} - 2000 \, a^{4} b c d^{3} + 625 \, a^{5} d^{4}\right )} e^{6}}\right ) + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (4 \, b c - 5 \, a d\right )} \sqrt {e x} e + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (4 \, b c - 5 \, a d\right )} \sqrt {e x} e - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {{\left (256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}\right )} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right )}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

1/8*(4*(b*d*e*x^2 - (4*b*c - 5*a*d)*e)*(b*x^2 + a)^(3/4)*sqrt(e*x) + 4*(b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*
a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4)*arctan(((4*b^8*c - 5*a*b^7
*d)*(b*x^2 + a)^(3/4)*sqrt(e*x)*e*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 +
625*a^4*d^4)*e^6/b^9)^(3/4) + (b^8*x^2 + a*b^7)*sqrt(((16*b^2*c^2 - 40*a*b*c*d + 25*a^2*d^2)*sqrt(b*x^2 + a)*
e^3*x + (b^5*x^2 + a*b^4)*sqrt((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625
*a^4*d^4)*e^6/b^9))/(b*x^2 + a))*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 +
625*a^4*d^4)*e^6/b^9)^(3/4))/((256*b^5*c^4 - 1280*a*b^4*c^3*d + 2400*a^2*b^3*c^2*d^2 - 2000*a^3*b^2*c*d^3 + 62
5*a^4*b*d^4)*e^6*x^2 + (256*a*b^4*c^4 - 1280*a^2*b^3*c^3*d + 2400*a^3*b^2*c^2*d^2 - 2000*a^4*b*c*d^3 + 625*a^5
*d^4)*e^6)) + (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 6
25*a^4*d^4)*e^6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(e*x)*e + (b^3*x^2 + a*b^2)*((256*b^4*c
^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4))/(b*x^2 + a)) -
(b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^
6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(e*x)*e - (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^
3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4))/(b*x^2 + a)))/(b^3*x^2 + a*b^
2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(5/4), x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

[Out]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^{3/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x)

[Out]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(5/4), x)

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sympy [C]  time = 32.83, size = 94, normalized size = 0.55 \[ \frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(d*x**2+c)/(b*x**2+a)**(5/4),x)

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(9/4)) + d
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(13/4))

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