3.7.84 \(\int \frac {(c+d x^2)^{5/2}}{x^4 (a+b x^2)} \, dx\)
Optimal. Leaf size=130 \[ \frac {(b c-a d)^{5/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} b}+\frac {c \sqrt {c+d x^2} (b c-2 a d)}{a^2 x}-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b} \]
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Rubi [A] time = 0.18, antiderivative size = 130, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used =
{474, 580, 523, 217, 206, 377, 205} \begin {gather*} \frac {c \sqrt {c+d x^2} (b c-2 a d)}{a^2 x}+\frac {(b c-a d)^{5/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} b}-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)),x]
[Out]
(c*(b*c - 2*a*d)*Sqrt[c + d*x^2])/(a^2*x) - (c*(c + d*x^2)^(3/2))/(3*a*x^3) + ((b*c - a*d)^(5/2)*ArcTan[(Sqrt[
b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(5/2)*b) + (d^(5/2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/b
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 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 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 474
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)
*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 580
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] && !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])
Rubi steps
\begin {align*} \int \frac {\left (c+d x^2\right )^{5/2}}{x^4 \left (a+b x^2\right )} \, dx &=-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {\int \frac {\sqrt {c+d x^2} \left (-3 c (b c-2 a d)+3 a d^2 x^2\right )}{x^2 \left (a+b x^2\right )} \, dx}{3 a}\\ &=\frac {c (b c-2 a d) \sqrt {c+d x^2}}{a^2 x}-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {\int \frac {3 c \left (b^2 c^2-3 a b c d+3 a^2 d^2\right )+3 a^2 d^3 x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{3 a^2}\\ &=\frac {c (b c-2 a d) \sqrt {c+d x^2}}{a^2 x}-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {d^3 \int \frac {1}{\sqrt {c+d x^2}} \, dx}{b}+\frac {(b c-a d)^3 \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{a^2 b}\\ &=\frac {c (b c-2 a d) \sqrt {c+d x^2}}{a^2 x}-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b}+\frac {(b c-a d)^3 \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{a^2 b}\\ &=\frac {c (b c-2 a d) \sqrt {c+d x^2}}{a^2 x}-\frac {c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac {(b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} b}+\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 125, normalized size = 0.96 \begin {gather*} \frac {(b c-a d)^{5/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} b}+\frac {c \sqrt {c+d x^2} \left (3 b c x^2-a \left (c+7 d x^2\right )\right )}{3 a^2 x^3}+\frac {d^{5/2} \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{b} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)),x]
[Out]
(c*Sqrt[c + d*x^2]*(3*b*c*x^2 - a*(c + 7*d*x^2)))/(3*a^2*x^3) + ((b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/
(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(5/2)*b) + (d^(5/2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/b
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IntegrateAlgebraic [A] time = 0.49, size = 172, normalized size = 1.32 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-a c^2-7 a c d x^2+3 b c^2 x^2\right )}{3 a^2 x^3}-\frac {\sqrt {b c-a d} \left (a^2 d^2-2 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac {a \sqrt {d}-b x \sqrt {c+d x^2}+b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{5/2} b}-\frac {d^{5/2} \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{b} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)),x]
[Out]
(Sqrt[c + d*x^2]*(-(a*c^2) + 3*b*c^2*x^2 - 7*a*c*d*x^2))/(3*a^2*x^3) - (Sqrt[b*c - a*d]*(b^2*c^2 - 2*a*b*c*d +
a^2*d^2)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^2 - b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sqrt[b*c - a*d])])/(a^(5/2)*b) - (
d^(5/2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/b
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fricas [A] time = 2.05, size = 901, normalized size = 6.93 \begin {gather*} \left [\frac {6 \, a^{2} d^{\frac {5}{2}} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{3} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (a b c^{2} - {\left (3 \, b^{2} c^{2} - 7 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, a^{2} b x^{3}}, -\frac {12 \, a^{2} \sqrt {-d} d^{2} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{3} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (a b c^{2} - {\left (3 \, b^{2} c^{2} - 7 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, a^{2} b x^{3}}, \frac {3 \, a^{2} d^{\frac {5}{2}} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{3} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) - 2 \, {\left (a b c^{2} - {\left (3 \, b^{2} c^{2} - 7 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, a^{2} b x^{3}}, -\frac {6 \, a^{2} \sqrt {-d} d^{2} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{3} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, {\left (a b c^{2} - {\left (3 \, b^{2} c^{2} - 7 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, a^{2} b x^{3}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a),x, algorithm="fricas")
[Out]
[1/12*(6*a^2*d^(5/2)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x
^3*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 -
4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(a
*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(a^2*b*x^3), -1/12*(12*a^2*sqrt(-d)*d^2*x^3*arctan(sqrt
(-d)*x/sqrt(d*x^2 + c)) - 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^3*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d
+ 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 +
c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(a*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2
+ c))/(a^2*b*x^3), 1/6*(3*a^2*d^(5/2)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 3*(b^2*c^2 - 2*a*b
*c*d + a^2*d^2)*x^3*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/
a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) - 2*(a*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(a^
2*b*x^3), -1/6*(6*a^2*sqrt(-d)*d^2*x^3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*
x^3*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d
^2)*x^3 + (b*c^2 - a*c*d)*x)) + 2*(a*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(a^2*b*x^3)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:
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maple [B] time = 0.02, size = 3346, normalized size = 25.74 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x^2+c)^(5/2)/x^4/(b*x^2+a),x)
[Out]
3/2*b/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c
)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b
))*d*c^2-3/2*b/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*
(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*
b)^(1/2)/b))*d*c^2+1/2*b^2/a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*
(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^
(1/2))/(x+(-a*b)^(1/2)/b))*c^3+7/16*b/a^2*d*c*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a
*d-b*c)/b)^(1/2)*x-15/8/a^2*b*d*c*x*(d*x^2+c)^(1/2)-1/2*b^2/a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)
^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(
-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*c^3-1/a^2*b*d/c*x*(d*x^2+c)^(5/2)+7/16*b/a^2*d*c*((
x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x+b/a/(-a*b)^(1/2)*((x+(-a*b)^(
1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d*c-1/2/b*a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(
1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2
*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*d^3-b/a/(-a*b)^(1/2)*((x-(-a*b)^(
1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d*c+1/2/b*a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(
1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*
(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*d^3-5/4/a*d^(3/2)*ln(((x+(-a*b)^(1
/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^
(1/2))*c-1/4/a*d^2*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-1/10*b^2
/a^2/(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(5/2)-1/3/a/c/x^3
*(d*x^2+c)^(7/2)+5/2/a*d^2*x*(d*x^2+c)^(1/2)+5/2/a*d^(3/2)*c*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/10*b^2/a^2/(-a*b)
^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(5/2)-1/4/a*d^2*((x-(-a*b)^(
1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-5/4/a*d^(3/2)*ln(((x-(-a*b)^(1/2)/b)*d+
(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c-
15/8/a^2*b*d^(1/2)*c^2*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-4/3/a*d/c^2/x*(d*x^2+c)^(7/2)+4/3/a*d^2/c^2*x*(d*x^2+c)^(
5/2)+5/3/a*d^2/c*x*(d*x^2+c)^(3/2)+3/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b
)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-
b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*d^2*c+1/8*b/a^2*d*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)
/b*d-(a*d-b*c)/b)^(3/2)*x+15/16*b/a^2*d^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1
/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c^2-1/6*b/a/(-a*b)^(1/2)*((x-(-a*b)^(1/2)
/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*d+1/6*b^2/a^2/(-a*b)^(1/2)*((x-(-a*b)^(1/2)/b
)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*c+1/2*b^2/a^2/(-a*b)^(1/2)*((x-(-a*b)^(1/2)/b)^
2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c^2-3/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(
-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)
*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*d^2*c+1/8*b/a^2*d*((x+(-a*b)^(1/2)/b)^2*d-2*(-
a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x+15/16*b/a^2*d^(1/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/
2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c^2+1/6*b/a/
(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*d-1/6*b^2/a^2/(-
a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*c-1/2*b^2/a^2/(-a*
b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c^2+1/a^2*b/c/x*(d*x
^2+c)^(7/2)-5/4/a^2*b*d*x*(d*x^2+c)^(3/2)-1/2/(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1
/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d^2+1/2/b*d^(5/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^
(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))+1/2/(-a*b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*
d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d^2+1/2/b*d^(5/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(
1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))
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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 )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )} x^{4}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a),x, algorithm="maxima")
[Out]
integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)*x^4), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^{5/2}}{x^4\,\left (b\,x^2+a\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c + d*x^2)^(5/2)/(x^4*(a + b*x^2)),x)
[Out]
int((c + d*x^2)^(5/2)/(x^4*(a + b*x^2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{x^{4} \left (a + b x^{2}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x**2+c)**(5/2)/x**4/(b*x**2+a),x)
[Out]
Integral((c + d*x**2)**(5/2)/(x**4*(a + b*x**2)), x)
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