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

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

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Rubi [A]  time = 0.16, antiderivative size = 191, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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 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 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 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])

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*}

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Mathematica [A]  time = 0.24, size = 167, normalized size = 0.87 \begin {gather*} \frac {8 a^4 e^5+a^3 c e^3 \left (-20 d^2-15 d e x+12 e^2 x^2\right )+a^2 c^2 e \left (-5 d^4-30 d^2 e^2 x^2-20 d e^3 x^3+3 e^4 x^4\right )+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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 1.04, size = 192, normalized size = 1.01 \begin {gather*} \frac {8 a^4 e^5-20 a^3 c d^2 e^3-15 a^3 c d e^4 x+12 a^3 c e^5 x^2-5 a^2 c^2 d^4 e-30 a^2 c^2 d^2 e^3 x^2-20 a^2 c^2 d e^4 x^3+3 a^2 c^2 e^5 x^4+3 a c^3 d^5 x+10 a c^3 d^3 e^2 x^3+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 {a+c x^2}-\sqrt {c} x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 488, normalized size = 2.55 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.27, size = 199, normalized size = 1.04 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.07, size = 270, normalized size = 1.41 \begin {gather*} \frac {e^{5} x^{4}}{\left (c \,x^{2}+a \right )^{\frac {3}{2}} c}-\frac {5 d \,e^{4} x^{3}}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c}+\frac {4 a \,e^{5} x^{2}}{\left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{2}}-\frac {10 d^{2} e^{3} x^{2}}{\left (c \,x^{2}+a \right )^{\frac {3}{2}} c}+\frac {d^{5} x}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a}-\frac {10 d^{3} e^{2} x}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c}+\frac {8 a^{2} e^{5}}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{3}}-\frac {20 a \,d^{2} e^{3}}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{2}}+\frac {10 d^{3} e^{2} x}{3 \sqrt {c \,x^{2}+a}\, a c}+\frac {2 d^{5} x}{3 \sqrt {c \,x^{2}+a}\, a^{2}}-\frac {5 d^{4} e}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c}-\frac {5 d \,e^{4} x}{\sqrt {c \,x^{2}+a}\, c^{2}}+\frac {5 d \,e^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {5}{2}}} \end {gather*}

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(c^(1/2)*x+(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 [A]  time = 1.43, size = 281, normalized size = 1.47 \begin {gather*} -\frac {5}{3} \, d e^{4} x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {2 \, a}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}}\right )} + \frac {e^{5} x^{4}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} - \frac {10 \, d^{2} e^{3} x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {4 \, a e^{5} x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}} + \frac {2 \, d^{5} x}{3 \, \sqrt {c x^{2} + a} a^{2}} + \frac {d^{5} x}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {10 \, d^{3} e^{2} x}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {10 \, d^{3} e^{2} x}{3 \, \sqrt {c x^{2} + a} a c} - \frac {5 \, d e^{4} x}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {5 \, d e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {5}{2}}} - \frac {5 \, d^{4} e}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c} - \frac {20 \, a d^{2} e^{3}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}} + \frac {8 \, a^{2} e^{5}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{3}} \end {gather*}

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^5/(a + c*x^2)^(5/2),x)

[Out]

int((d + e*x)^5/(a + c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{5}}{\left (a + c x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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