### 3.577 $$\int \frac{(d+e x)^5}{(a+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=191 $-\frac{e \sqrt{a+c x^2} \left (4 \left (-2 a^2 e^4+4 a c d^2 e^2+c^2 d^4\right )+c d e x \left (7 a e^2+2 c d^2\right )\right )}{3 a^2 c^3}-\frac{2 (d+e x)^2 \left (2 a^2 e^3-c d x \left (3 a e^2+c d^2\right )\right )}{3 a^2 c^2 \sqrt{a+c x^2}}+\frac{5 d e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{5/2}}-\frac{(d+e x)^4 (a e-c d x)}{3 a c \left (a+c x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.163214, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.263, Rules used = {739, 819, 780, 217, 206} $-\frac{e \sqrt{a+c x^2} \left (4 \left (-2 a^2 e^4+4 a c d^2 e^2+c^2 d^4\right )+c d e x \left (7 a e^2+2 c d^2\right )\right )}{3 a^2 c^3}-\frac{2 (d+e x)^2 \left (2 a^2 e^3-c d x \left (3 a e^2+c d^2\right )\right )}{3 a^2 c^2 \sqrt{a+c x^2}}+\frac{5 d e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{5/2}}-\frac{(d+e x)^4 (a e-c d x)}{3 a c \left (a+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^5/(a + c*x^2)^(5/2),x]

[Out]

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

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.245504, size = 167, normalized size = 0.87 $\frac{a^2 c^2 e \left (-30 d^2 e^2 x^2-5 d^4-20 d e^3 x^3+3 e^4 x^4\right )+a^3 c e^3 \left (-20 d^2-15 d e x+12 e^2 x^2\right )+8 a^4 e^5+a c^3 d^3 x \left (3 d^2+10 e^2 x^2\right )+2 c^4 d^5 x^3}{3 a^2 c^3 \left (a+c x^2\right )^{3/2}}+\frac{5 d e^4 \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(8*a^4*e^5 + 2*c^4*d^5*x^3 + a*c^3*d^3*x*(3*d^2 + 10*e^2*x^2) + a^3*c*e^3*(-20*d^2 - 15*d*e*x + 12*e^2*x^2) +
a^2*c^2*e*(-5*d^4 - 30*d^2*e^2*x^2 - 20*d*e^3*x^3 + 3*e^4*x^4))/(3*a^2*c^3*(a + c*x^2)^(3/2)) + (5*d*e^4*Log[c
*x + Sqrt[c]*Sqrt[a + c*x^2]])/c^(5/2)

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Maple [A]  time = 0.053, size = 270, normalized size = 1.4 \begin{align*}{\frac{{e}^{5}{x}^{4}}{c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+4\,{\frac{a{e}^{5}{x}^{2}}{{c}^{2} \left ( c{x}^{2}+a \right ) ^{3/2}}}+{\frac{8\,{e}^{5}{a}^{2}}{3\,{c}^{3}} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,d{e}^{4}{x}^{3}}{3\,c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-5\,{\frac{d{e}^{4}x}{{c}^{2}\sqrt{c{x}^{2}+a}}}+5\,{\frac{d{e}^{4}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) }{{c}^{5/2}}}-10\,{\frac{{d}^{2}{e}^{3}{x}^{2}}{c \left ( c{x}^{2}+a \right ) ^{3/2}}}-{\frac{20\,{d}^{2}{e}^{3}a}{3\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{10\,{d}^{3}{e}^{2}x}{3\,c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{10\,{d}^{3}{e}^{2}x}{3\,ac}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{5\,{d}^{4}e}{3\,c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{{d}^{5}x}{3\,a} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{d}^{5}x}{3\,{a}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^5*x^4/c/(c*x^2+a)^(3/2)+4*e^5*a/c^2*x^2/(c*x^2+a)^(3/2)+8/3*e^5*a^2/c^3/(c*x^2+a)^(3/2)-5/3*d*e^4*x^3/c/(c*x
^2+a)^(3/2)-5*d*e^4/c^2*x/(c*x^2+a)^(1/2)+5*d*e^4/c^(5/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))-10*d^2*e^3*x^2/c/(c*x^
2+a)^(3/2)-20/3*d^2*e^3*a/c^2/(c*x^2+a)^(3/2)-10/3*d^3*e^2/c*x/(c*x^2+a)^(3/2)+10/3*d^3*e^2/a/c*x/(c*x^2+a)^(1
/2)-5/3*d^4*e/c/(c*x^2+a)^(3/2)+1/3*d^5*x/a/(c*x^2+a)^(3/2)+2/3*d^5/a^2*x/(c*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.4707, size = 1008, normalized size = 5.28 \begin{align*} \left [\frac{15 \,{\left (a^{2} c^{2} d e^{4} x^{4} + 2 \, a^{3} c d e^{4} x^{2} + a^{4} d e^{4}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (3 \, a^{2} c^{2} e^{5} x^{4} - 5 \, a^{2} c^{2} d^{4} e - 20 \, a^{3} c d^{2} e^{3} + 8 \, a^{4} e^{5} + 2 \,{\left (c^{4} d^{5} + 5 \, a c^{3} d^{3} e^{2} - 10 \, a^{2} c^{2} d e^{4}\right )} x^{3} - 6 \,{\left (5 \, a^{2} c^{2} d^{2} e^{3} - 2 \, a^{3} c e^{5}\right )} x^{2} + 3 \,{\left (a c^{3} d^{5} - 5 \, a^{3} c d e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{6 \,{\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}}, -\frac{15 \,{\left (a^{2} c^{2} d e^{4} x^{4} + 2 \, a^{3} c d e^{4} x^{2} + a^{4} d e^{4}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (3 \, a^{2} c^{2} e^{5} x^{4} - 5 \, a^{2} c^{2} d^{4} e - 20 \, a^{3} c d^{2} e^{3} + 8 \, a^{4} e^{5} + 2 \,{\left (c^{4} d^{5} + 5 \, a c^{3} d^{3} e^{2} - 10 \, a^{2} c^{2} d e^{4}\right )} x^{3} - 6 \,{\left (5 \, a^{2} c^{2} d^{2} e^{3} - 2 \, a^{3} c e^{5}\right )} x^{2} + 3 \,{\left (a c^{3} d^{5} - 5 \, a^{3} c d e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{3 \,{\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a^2*c^2*d*e^4*x^4 + 2*a^3*c*d*e^4*x^2 + a^4*d*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*
x - a) + 2*(3*a^2*c^2*e^5*x^4 - 5*a^2*c^2*d^4*e - 20*a^3*c*d^2*e^3 + 8*a^4*e^5 + 2*(c^4*d^5 + 5*a*c^3*d^3*e^2
- 10*a^2*c^2*d*e^4)*x^3 - 6*(5*a^2*c^2*d^2*e^3 - 2*a^3*c*e^5)*x^2 + 3*(a*c^3*d^5 - 5*a^3*c*d*e^4)*x)*sqrt(c*x^
2 + a))/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3), -1/3*(15*(a^2*c^2*d*e^4*x^4 + 2*a^3*c*d*e^4*x^2 + a^4*d*e^4)*
sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (3*a^2*c^2*e^5*x^4 - 5*a^2*c^2*d^4*e - 20*a^3*c*d^2*e^3 + 8*a^4*
e^5 + 2*(c^4*d^5 + 5*a*c^3*d^3*e^2 - 10*a^2*c^2*d*e^4)*x^3 - 6*(5*a^2*c^2*d^2*e^3 - 2*a^3*c*e^5)*x^2 + 3*(a*c^
3*d^5 - 5*a^3*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{5}}{\left (a + c x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**5/(a + c*x**2)**(5/2), x)

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Giac [A]  time = 1.3285, size = 269, normalized size = 1.41 \begin{align*} -\frac{5 \, d e^{4} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{c^{\frac{5}{2}}} + \frac{{\left ({\left (x{\left (\frac{3 \, x e^{5}}{c} + \frac{2 \,{\left (c^{6} d^{5} + 5 \, a c^{5} d^{3} e^{2} - 10 \, a^{2} c^{4} d e^{4}\right )}}{a^{2} c^{5}}\right )} - \frac{6 \,{\left (5 \, a^{2} c^{4} d^{2} e^{3} - 2 \, a^{3} c^{3} e^{5}\right )}}{a^{2} c^{5}}\right )} x + \frac{3 \,{\left (a c^{5} d^{5} - 5 \, a^{3} c^{3} d e^{4}\right )}}{a^{2} c^{5}}\right )} x - \frac{5 \, a^{2} c^{4} d^{4} e + 20 \, a^{3} c^{3} d^{2} e^{3} - 8 \, a^{4} c^{2} e^{5}}{a^{2} c^{5}}}{3 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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