3.1459 \(\int \frac{(A+B x) (d+e x)^{5/2}}{(a-c x^2)^3} \, dx\)
Optimal. Leaf size=372 \[ \frac{\sqrt{\sqrt{c} d-\sqrt{a} e} \left (5 a B e \left (\sqrt{a} e+2 \sqrt{c} d\right )-3 A \left (2 \sqrt{a} c d e-a \sqrt{c} e^2+4 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{9/4}}-\frac{\sqrt{\sqrt{a} e+\sqrt{c} d} \left (5 a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-A \left (-6 \sqrt{a} c d e-3 a \sqrt{c} e^2+12 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{9/4}}+\frac{\sqrt{d+e x} \left (c x \left (6 A c d^2-a e (3 A e+5 B d)\right )+a e (3 A c d-5 a B e)\right )}{16 a^2 c^2 \left (a-c x^2\right )}+\frac{(d+e x)^{3/2} (x (a B e+A c d)+a (A e+B d))}{4 a c \left (a-c x^2\right )^2} \]
[Out]
((d + e*x)^(3/2)*(a*(B*d + A*e) + (A*c*d + a*B*e)*x))/(4*a*c*(a - c*x^2)^2) + (Sqrt[d + e*x]*(a*e*(3*A*c*d - 5
*a*B*e) + c*(6*A*c*d^2 - a*e*(5*B*d + 3*A*e))*x))/(16*a^2*c^2*(a - c*x^2)) + (Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(5*a
*B*e*(2*Sqrt[c]*d + Sqrt[a]*e) - 3*A*(4*c^(3/2)*d^2 + 2*Sqrt[a]*c*d*e - a*Sqrt[c]*e^2))*ArcTanh[(c^(1/4)*Sqrt[
d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(9/4)) - (Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(5*a*B*e*(2*Sqrt[c
]*d - Sqrt[a]*e) - A*(12*c^(3/2)*d^2 - 6*Sqrt[a]*c*d*e - 3*a*Sqrt[c]*e^2))*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqr
t[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(9/4))
________________________________________________________________________________________
Rubi [A] time = 0.686117, antiderivative size = 372, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{819, 821, 827, 1166, 208} \[ \frac{\sqrt{\sqrt{c} d-\sqrt{a} e} \left (5 a B e \left (\sqrt{a} e+2 \sqrt{c} d\right )-3 A \left (2 \sqrt{a} c d e-a \sqrt{c} e^2+4 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{9/4}}-\frac{\sqrt{\sqrt{a} e+\sqrt{c} d} \left (5 a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-A \left (-6 \sqrt{a} c d e-3 a \sqrt{c} e^2+12 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{9/4}}+\frac{\sqrt{d+e x} \left (c x \left (6 A c d^2-a e (3 A e+5 B d)\right )+a e (3 A c d-5 a B e)\right )}{16 a^2 c^2 \left (a-c x^2\right )}+\frac{(d+e x)^{3/2} (x (a B e+A c d)+a (A e+B d))}{4 a c \left (a-c x^2\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*(d + e*x)^(5/2))/(a - c*x^2)^3,x]
[Out]
((d + e*x)^(3/2)*(a*(B*d + A*e) + (A*c*d + a*B*e)*x))/(4*a*c*(a - c*x^2)^2) + (Sqrt[d + e*x]*(a*e*(3*A*c*d - 5
*a*B*e) + c*(6*A*c*d^2 - a*e*(5*B*d + 3*A*e))*x))/(16*a^2*c^2*(a - c*x^2)) + (Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(5*a
*B*e*(2*Sqrt[c]*d + Sqrt[a]*e) - 3*A*(4*c^(3/2)*d^2 + 2*Sqrt[a]*c*d*e - a*Sqrt[c]*e^2))*ArcTanh[(c^(1/4)*Sqrt[
d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(9/4)) - (Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(5*a*B*e*(2*Sqrt[c
]*d - Sqrt[a]*e) - A*(12*c^(3/2)*d^2 - 6*Sqrt[a]*c*d*e - 3*a*Sqrt[c]*e^2))*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqr
t[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(9/4))
Rule 819
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) || !ILtQ[m + 2*p + 3, 0])
Rule 821
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Rule 827
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]
Rule 1166
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]
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]
Rubi steps
\begin{align*} \int \frac{(A+B x) (d+e x)^{5/2}}{\left (a-c x^2\right )^3} \, dx &=\frac{(d+e x)^{3/2} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}-\frac{\int \frac{\sqrt{d+e x} \left (\frac{1}{2} \left (-6 A c d^2+a e (5 B d+3 A e)\right )-\frac{1}{2} e (3 A c d-5 a B e) x\right )}{\left (a-c x^2\right )^2} \, dx}{4 a c}\\ &=\frac{(d+e x)^{3/2} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}+\frac{\sqrt{d+e x} \left (a e (3 A c d-5 a B e)+c \left (6 A c d^2-a e (5 B d+3 A e)\right ) x\right )}{16 a^2 c^2 \left (a-c x^2\right )}+\frac{\int \frac{\frac{1}{4} \left (3 A c d \left (4 c d^2-3 a e^2\right )-5 a B e \left (2 c d^2-a e^2\right )\right )+\frac{1}{4} c e \left (6 A c d^2-a e (5 B d+3 A e)\right ) x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{8 a^2 c^2}\\ &=\frac{(d+e x)^{3/2} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}+\frac{\sqrt{d+e x} \left (a e (3 A c d-5 a B e)+c \left (6 A c d^2-a e (5 B d+3 A e)\right ) x\right )}{16 a^2 c^2 \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} c d e \left (6 A c d^2-a e (5 B d+3 A e)\right )+\frac{1}{4} e \left (3 A c d \left (4 c d^2-3 a e^2\right )-5 a B e \left (2 c d^2-a e^2\right )\right )+\frac{1}{4} c e \left (6 A c d^2-a e (5 B d+3 A e)\right ) x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt{d+e x}\right )}{4 a^2 c^2}\\ &=\frac{(d+e x)^{3/2} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}+\frac{\sqrt{d+e x} \left (a e (3 A c d-5 a B e)+c \left (6 A c d^2-a e (5 B d+3 A e)\right ) x\right )}{16 a^2 c^2 \left (a-c x^2\right )}+\frac{\left (\left (\sqrt{c} d-\sqrt{a} e\right ) \left (5 a B e \left (2 \sqrt{c} d+\sqrt{a} e\right )-3 A \left (4 c^{3/2} d^2+2 \sqrt{a} c d e-a \sqrt{c} e^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d-\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{32 a^{5/2} c^{3/2}}+\frac{\left (\frac{1}{8} c e \left (6 A c d^2-a e (5 B d+3 A e)\right )-\frac{-\frac{1}{2} c^2 d e \left (6 A c d^2-a e (5 B d+3 A e)\right )-2 c \left (-\frac{1}{4} c d e \left (6 A c d^2-a e (5 B d+3 A e)\right )+\frac{1}{4} e \left (3 A c d \left (4 c d^2-3 a e^2\right )-5 a B e \left (2 c d^2-a e^2\right )\right )\right )}{4 \sqrt{a} \sqrt{c} e}\right ) \operatorname{Subst}\left (\int \frac{1}{c d+\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{4 a^2 c^2}\\ &=\frac{(d+e x)^{3/2} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}+\frac{\sqrt{d+e x} \left (a e (3 A c d-5 a B e)+c \left (6 A c d^2-a e (5 B d+3 A e)\right ) x\right )}{16 a^2 c^2 \left (a-c x^2\right )}+\frac{\sqrt{\sqrt{c} d-\sqrt{a} e} \left (5 a B e \left (2 \sqrt{c} d+\sqrt{a} e\right )-3 A \left (4 c^{3/2} d^2+2 \sqrt{a} c d e-a \sqrt{c} e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{9/4}}-\frac{\sqrt{\sqrt{c} d+\sqrt{a} e} \left (5 a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-3 A \left (4 c^{3/2} d^2-2 \sqrt{a} c d e-a \sqrt{c} e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d+\sqrt{a} e}}\right )}{32 a^{5/2} c^{9/4}}\\ \end{align*}
Mathematica [A] time = 1.68072, size = 569, normalized size = 1.53 \[ \frac{\frac{c^2 (d+e x)^{7/2} \left (a^2 e^2 (A e+6 B d-B e x)-a c d^2 e (7 A+5 B x)+6 A c^2 d^3 x\right )}{2 \left (a-c x^2\right )}+\frac{\sqrt [4]{c} \left (5 a B d e \left (5 a e^2+7 c d^2\right )-3 A \left (-a^2 e^4+7 a c d^2 e^2+14 c^2 d^4\right )\right ) \left (2 \sqrt{a} c^{3/4} e \sqrt{d+e x} (7 d+e x)+3 \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-3 \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{12 \sqrt{a}}+\frac{2 a c^2 (d+e x)^{7/2} \left (c d^2-a e^2\right ) (-a A e+a B (d-e x)+A c d x)}{\left (a-c x^2\right )^2}+\frac{\left (6 A c^2 d^3-a B e \left (a e^2+5 c d^2\right )\right ) \left (2 \sqrt{a} \sqrt [4]{c} e \sqrt{d+e x} \left (15 a e^2+c \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )+15 \left (\sqrt{c} d-\sqrt{a} e\right )^{7/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-15 \left (\sqrt{a} e+\sqrt{c} d\right )^{7/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{12 \sqrt{a} \sqrt [4]{c}}}{8 a^2 c^2 \left (c d^2-a e^2\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*(d + e*x)^(5/2))/(a - c*x^2)^3,x]
[Out]
((2*a*c^2*(c*d^2 - a*e^2)*(d + e*x)^(7/2)*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(a - c*x^2)^2 + (c^2*(d + e*x)
^(7/2)*(6*A*c^2*d^3*x - a*c*d^2*e*(7*A + 5*B*x) + a^2*e^2*(6*B*d + A*e - B*e*x)))/(2*(a - c*x^2)) + (c^(1/4)*(
5*a*B*d*e*(7*c*d^2 + 5*a*e^2) - 3*A*(14*c^2*d^4 + 7*a*c*d^2*e^2 - a^2*e^4))*(2*Sqrt[a]*c^(3/4)*e*Sqrt[d + e*x]
*(7*d + e*x) + 3*(Sqrt[c]*d - Sqrt[a]*e)^(5/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] -
3*(Sqrt[c]*d + Sqrt[a]*e)^(5/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]))/(12*Sqrt[a]) +
((6*A*c^2*d^3 - a*B*e*(5*c*d^2 + a*e^2))*(2*Sqrt[a]*c^(1/4)*e*Sqrt[d + e*x]*(15*a*e^2 + c*(58*d^2 + 16*d*e*x +
3*e^2*x^2)) + 15*(Sqrt[c]*d - Sqrt[a]*e)^(7/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] -
15*(Sqrt[c]*d + Sqrt[a]*e)^(7/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]))/(12*Sqrt[a]*c
^(1/4)))/(8*a^2*c^2*(c*d^2 - a*e^2)^2)
________________________________________________________________________________________
Maple [B] time = 0.049, size = 1447, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*(e*x+d)^(5/2)/(-c*x^2+a)^3,x)
[Out]
-5/32*e^2/a/c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B*d+5/3
2*e^2/a/c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B*d-1/2*e^
3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(5/2)*A*d-15/16*e^2/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(5/2)*B*d^2+17/16*e^3/(c*e^2
*x^2-a*e^2)^2/a*(e*x+d)^(3/2)*A*d^2+15/16*e^2/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(3/2)*B*d^3+3/8*e^5/(c*e^2*x^2-a*e
^2)^2/c*(e*x+d)^(1/2)*A*d+5/16*e^2/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(7/2)*B*d-5/16*e^6/(c*e^2*x^2-a*e^2)^2*a/c^2*
(e*x+d)^(1/2)*B-5/16*e^2/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(1/2)*B*d^4-3/16*e/a^2/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)
*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d^2-3/4*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(1/2)*A*
d^3-3/32*e^3/a/c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A+3/
32*e^3/a/c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A-15/16*e
^4/(c*e^2*x^2-a*e^2)^2/c*(e*x+d)^(3/2)*B*d+5/8*e^4/(c*e^2*x^2-a*e^2)^2/c*(e*x+d)^(1/2)*B*d^2+5/32*e^4/c/(a*c*e
^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B+3/16*e/a^
2/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d^2+5/32*e^4/c/(a
*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B+1/16
*e^5/(c*e^2*x^2-a*e^2)^2/c*(e*x+d)^(3/2)*A+3/16*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(7/2)*A+9/16*e^4/(c*e^2*x^2-
a*e^2)^2/c*(e*x+d)^(5/2)*B+3/8*e/a^2*c/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c
/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d^3+3/8*e/a^2*c/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e
*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d^3-3/8*e/(c*e^2*x^2-a*e^2)^2/a^2*(e*x+d)^(7/2)*A*c*d^2+9/8*
e/(c*e^2*x^2-a*e^2)^2/a^2*c*(e*x+d)^(5/2)*A*d^3-9/8*e/(c*e^2*x^2-a*e^2)^2/a^2*c*(e*x+d)^(3/2)*A*d^4+3/8*e/(c*e
^2*x^2-a*e^2)^2/a^2*c*(e*x+d)^(1/2)*A*d^5-9/32*e^3/a/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((
e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d-5/16*e^2/a/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)
*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B*d^2-9/32*e^3/a/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1
/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d-5/16*e^2/a/(a*c*e^2)^(1/2)/((-c*d+(
a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B*d^2
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (c x^{2} - a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="maxima")
[Out]
-integrate((B*x + A)*(e*x + d)^(5/2)/(c*x^2 - a)^3, x)
________________________________________________________________________________________
Fricas [B] time = 7.9166, size = 7255, normalized size = 19.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="fricas")
[Out]
-1/64*((a^2*c^4*x^4 - 2*a^3*c^3*x^2 + a^4*c^2)*sqrt((144*A^2*c^3*d^5 - 240*A*B*a*c^2*d^4*e + 240*A*B*a^2*c*d^2
*e^3 - 30*A*B*a^3*e^5 + a^5*c^4*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B
^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a
^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))*log(-(4320*A^3*B*c^4*d^5*e^4 - 432*(25*A^2*B^2*a*c^3 + 3*A^4*c^4)*d^4*e^5
+ 360*(25*A*B^3*a^2*c^2 - 3*A^3*B*a*c^3)*d^3*e^6 - 4*(625*B^4*a^3*c - 1350*A^2*B^2*a^2*c^2 - 243*A^4*a*c^3)*d^
2*e^7 - 30*(125*A*B^3*a^3*c + 27*A^3*B*a^2*c^2)*d*e^8 + (625*B^4*a^4 - 81*A^4*a^2*c^2)*e^9)*sqrt(e*x + d) + (1
80*A^2*B*a^3*c^4*d^2*e^5 - 6*(50*A*B^2*a^4*c^3 + 9*A^3*a^3*c^4)*d*e^6 + 5*(25*B^3*a^5*c^2 + 9*A^2*B*a^4*c^3)*e
^7 - (12*A*a^5*c^8*d^2 - 10*B*a^6*c^7*d*e - 3*A*a^6*c^7*e^2)*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c
+ 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)))*sqrt((144*A^2*c^3*d^5 -
240*A*B*a*c^2*d^4*e + 240*A*B*a^2*c*d^2*e^3 - 30*A*B*a^3*e^5 + a^5*c^4*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*
A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)) + 20*(5*B^2*a^2
*c - 9*A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))) - (a^2*c^4*x^4 - 2*a^3*c^3*x^2 + a
^4*c^2)*sqrt((144*A^2*c^3*d^5 - 240*A*B*a*c^2*d^4*e + 240*A*B*a^2*c*d^2*e^3 - 30*A*B*a^3*e^5 + a^5*c^4*sqrt((9
00*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*
e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))*log
(-(4320*A^3*B*c^4*d^5*e^4 - 432*(25*A^2*B^2*a*c^3 + 3*A^4*c^4)*d^4*e^5 + 360*(25*A*B^3*a^2*c^2 - 3*A^3*B*a*c^3
)*d^3*e^6 - 4*(625*B^4*a^3*c - 1350*A^2*B^2*a^2*c^2 - 243*A^4*a*c^3)*d^2*e^7 - 30*(125*A*B^3*a^3*c + 27*A^3*B*
a^2*c^2)*d*e^8 + (625*B^4*a^4 - 81*A^4*a^2*c^2)*e^9)*sqrt(e*x + d) - (180*A^2*B*a^3*c^4*d^2*e^5 - 6*(50*A*B^2*
a^4*c^3 + 9*A^3*a^3*c^4)*d*e^6 + 5*(25*B^3*a^5*c^2 + 9*A^2*B*a^4*c^3)*e^7 - (12*A*a^5*c^8*d^2 - 10*B*a^6*c^7*d
*e - 3*A*a^6*c^7*e^2)*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 4
50*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)))*sqrt((144*A^2*c^3*d^5 - 240*A*B*a*c^2*d^4*e + 240*A*B*a^2*c*d^2
*e^3 - 30*A*B*a^3*e^5 + a^5*c^4*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B
^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a
^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))) + (a^2*c^4*x^4 - 2*a^3*c^3*x^2 + a^4*c^2)*sqrt((144*A^2*c^3*d^5 - 240*A*B
*a*c^2*d^4*e + 240*A*B*a^2*c*d^2*e^3 - 30*A*B*a^3*e^5 - a^5*c^4*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a
*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*
A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))*log(-(4320*A^3*B*c^4*d^5*e^4 - 432*(25*A^2
*B^2*a*c^3 + 3*A^4*c^4)*d^4*e^5 + 360*(25*A*B^3*a^2*c^2 - 3*A^3*B*a*c^3)*d^3*e^6 - 4*(625*B^4*a^3*c - 1350*A^2
*B^2*a^2*c^2 - 243*A^4*a*c^3)*d^2*e^7 - 30*(125*A*B^3*a^3*c + 27*A^3*B*a^2*c^2)*d*e^8 + (625*B^4*a^4 - 81*A^4*
a^2*c^2)*e^9)*sqrt(e*x + d) + (180*A^2*B*a^3*c^4*d^2*e^5 - 6*(50*A*B^2*a^4*c^3 + 9*A^3*a^3*c^4)*d*e^6 + 5*(25*
B^3*a^5*c^2 + 9*A^2*B*a^4*c^3)*e^7 + (12*A*a^5*c^8*d^2 - 10*B*a^6*c^7*d*e - 3*A*a^6*c^7*e^2)*sqrt((900*A^2*B^2
*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5
*c^9)))*sqrt((144*A^2*c^3*d^5 - 240*A*B*a*c^2*d^4*e + 240*A*B*a^2*c*d^2*e^3 - 30*A*B*a^3*e^5 - a^5*c^4*sqrt((9
00*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*
e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))) -
(a^2*c^4*x^4 - 2*a^3*c^3*x^2 + a^4*c^2)*sqrt((144*A^2*c^3*d^5 - 240*A*B*a*c^2*d^4*e + 240*A*B*a^2*c*d^2*e^3 -
30*A*B*a^3*e^5 - a^5*c^4*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2
+ 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a^3 - 3*
A^2*a^2*c)*d*e^4)/(a^5*c^4))*log(-(4320*A^3*B*c^4*d^5*e^4 - 432*(25*A^2*B^2*a*c^3 + 3*A^4*c^4)*d^4*e^5 + 360*(
25*A*B^3*a^2*c^2 - 3*A^3*B*a*c^3)*d^3*e^6 - 4*(625*B^4*a^3*c - 1350*A^2*B^2*a^2*c^2 - 243*A^4*a*c^3)*d^2*e^7 -
30*(125*A*B^3*a^3*c + 27*A^3*B*a^2*c^2)*d*e^8 + (625*B^4*a^4 - 81*A^4*a^2*c^2)*e^9)*sqrt(e*x + d) - (180*A^2*
B*a^3*c^4*d^2*e^5 - 6*(50*A*B^2*a^4*c^3 + 9*A^3*a^3*c^4)*d*e^6 + 5*(25*B^3*a^5*c^2 + 9*A^2*B*a^4*c^3)*e^7 + (1
2*A*a^5*c^8*d^2 - 10*B*a^6*c^7*d*e - 3*A*a^6*c^7*e^2)*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a*c + 9*A^3
*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)))*sqrt((144*A^2*c^3*d^5 - 240*A*B
*a*c^2*d^4*e + 240*A*B*a^2*c*d^2*e^3 - 30*A*B*a^3*e^5 - a^5*c^4*sqrt((900*A^2*B^2*c^2*d^2*e^8 - 60*(25*A*B^3*a
*c + 9*A^3*B*c^2)*d*e^9 + (625*B^4*a^2 + 450*A^2*B^2*a*c + 81*A^4*c^2)*e^10)/(a^5*c^9)) + 20*(5*B^2*a^2*c - 9*
A^2*a*c^2)*d^3*e^2 - 15*(5*B^2*a^3 - 3*A^2*a^2*c)*d*e^4)/(a^5*c^4))) - 4*(4*B*a^2*c*d^2 + 7*A*a^2*c*d*e - 5*B*
a^3*e^2 - (6*A*c^3*d^2 - 5*B*a*c^2*d*e - 3*A*a*c^2*e^2)*x^3 + (A*a*c^2*d*e + 9*B*a^2*c*e^2)*x^2 + (10*A*a*c^2*
d^2 + 3*B*a^2*c*d*e + A*a^2*c*e^2)*x)*sqrt(e*x + d))/(a^2*c^4*x^4 - 2*a^3*c^3*x^2 + a^4*c^2)
________________________________________________________________________________________
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((B*x+A)*(e*x+d)**(5/2)/(-c*x**2+a)**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [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((B*x+A)*(e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="giac")
[Out]
Timed out