3.1.57 \(\int \frac {(a+b x^2)^{3/2}}{c+d x^2} \, dx\) [57]

Optimal. Leaf size=113 \[ \frac {b x \sqrt {a+b x^2}}{2 d}-\frac {\sqrt {b} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 d^2}+\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} d^2} \]

[Out]

-1/2*(-3*a*d+2*b*c)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)/d^2+(-a*d+b*c)^(3/2)*arctanh(x*(-a*d+b*c)^(1/2)
/c^(1/2)/(b*x^2+a)^(1/2))/d^2/c^(1/2)+1/2*b*x*(b*x^2+a)^(1/2)/d

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Rubi [A]
time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {427, 537, 223, 212, 385, 214} \begin {gather*} \frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} d^2}-\frac {\sqrt {b} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 d^2}+\frac {b x \sqrt {a+b x^2}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(c + d*x^2),x]

[Out]

(b*x*Sqrt[a + b*x^2])/(2*d) - (Sqrt[b]*(2*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*d^2) + ((b*c -
a*d)^(3/2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*d^2)

Rule 212

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

Rule 214

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

Rule 223

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 385

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 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 537

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]

Rubi steps

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

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Mathematica [A]
time = 0.31, size = 126, normalized size = 1.12 \begin {gather*} \frac {b d x \sqrt {a+b x^2}-\frac {2 (-b c+a d)^{3/2} \tan ^{-1}\left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {c}}+\sqrt {b} (2 b c-3 a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[a + b*x^2] - (2*(-(b*c) + a*d)^(3/2)*ArcTan[(-(d*x*Sqrt[a + b*x^2]) + Sqrt[b]*(c + d*x^2))/(Sqrt[c
]*Sqrt[-(b*c) + a*d])])/Sqrt[c] + Sqrt[b]*(2*b*c - 3*a*d)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*d^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1226\) vs. \(2(91)=182\).
time = 0.10, size = 1227, normalized size = 10.86 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

1/2/(-c*d)^(1/2)*(1/3*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)+b*(-c*d
)^(1/2)/d*(1/4*(2*b*(x-(-c*d)^(1/2)/d)+2*b*(-c*d)^(1/2)/d)/b*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c
*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)+1/8*(4*b*(a*d-b*c)/d+4*b^2*c/d)/b^(3/2)*ln((b*(-c*d)^(1/2)/d+b*(x-(-c*d)^(1/2)
/d))/b^(1/2)+((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)))+(a*d-b*c)/d*((
(x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)+b^(1/2)*(-c*d)^(1/2)/d*ln((b*(
-c*d)^(1/2)/d+b*(x-(-c*d)^(1/2)/d))/b^(1/2)+((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d
-b*c)/d)^(1/2))-(a*d-b*c)/d/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*((a*
d-b*c)/d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x-(-c*d)^(1
/2)/d))))-1/2/(-c*d)^(1/2)*(1/3*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/
2)-b*(-c*d)^(1/2)/d*(1/4*(2*b*(x+(-c*d)^(1/2)/d)-2*b*(-c*d)^(1/2)/d)/b*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2
)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)+1/8*(4*b*(a*d-b*c)/d+4*b^2*c/d)/b^(3/2)*ln((-b*(-c*d)^(1/2)/d+b*(x+(
-c*d)^(1/2)/d))/b^(1/2)+((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)))+(a*
d-b*c)/d*(((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)-b^(1/2)*(-c*d)^(1/2
)/d*ln((-b*(-c*d)^(1/2)/d+b*(x+(-c*d)^(1/2)/d))/b^(1/2)+((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(
1/2)/d)+(a*d-b*c)/d)^(1/2))-(a*d-b*c)/d/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/
2)/d)+2*((a*d-b*c)/d)^(1/2)*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/
(x+(-c*d)^(1/2)/d))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.79, size = 721, normalized size = 6.38 \begin {gather*} \left [\frac {2 \, \sqrt {b x^{2} + a} b d x - {\left (2 \, b c - 3 \, a d\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a c^{2} x + {\left (2 \, b c^{2} - a c d\right )} x^{3}\right )} \sqrt {b x^{2} + a} \sqrt {\frac {b c - a d}{c}}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right )}{4 \, d^{2}}, \frac {2 \, \sqrt {b x^{2} + a} b d x + 2 \, {\left (2 \, b c - 3 \, a d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a c^{2} x + {\left (2 \, b c^{2} - a c d\right )} x^{3}\right )} \sqrt {b x^{2} + a} \sqrt {\frac {b c - a d}{c}}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right )}{4 \, d^{2}}, \frac {2 \, \sqrt {b x^{2} + a} b d x - 2 \, {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (\frac {{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a} \sqrt {-\frac {b c - a d}{c}}}{2 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + {\left (a b c - a^{2} d\right )} x\right )}}\right ) - {\left (2 \, b c - 3 \, a d\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{4 \, d^{2}}, \frac {\sqrt {b x^{2} + a} b d x + {\left (2 \, b c - 3 \, a d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (\frac {{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a} \sqrt {-\frac {b c - a d}{c}}}{2 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + {\left (a b c - a^{2} d\right )} x\right )}}\right )}{2 \, d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(b*x^2 + a)*b*d*x - (2*b*c - 3*a*d)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - (b*c
- a*d)*sqrt((b*c - a*d)/c)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x
^2 - 4*(a*c^2*x + (2*b*c^2 - a*c*d)*x^3)*sqrt(b*x^2 + a)*sqrt((b*c - a*d)/c))/(d^2*x^4 + 2*c*d*x^2 + c^2)))/d^
2, 1/4*(2*sqrt(b*x^2 + a)*b*d*x + 2*(2*b*c - 3*a*d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (b*c - a*d)*
sqrt((b*c - a*d)/c)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 - 4*(
a*c^2*x + (2*b*c^2 - a*c*d)*x^3)*sqrt(b*x^2 + a)*sqrt((b*c - a*d)/c))/(d^2*x^4 + 2*c*d*x^2 + c^2)))/d^2, 1/4*(
2*sqrt(b*x^2 + a)*b*d*x - 2*(b*c - a*d)*sqrt(-(b*c - a*d)/c)*arctan(1/2*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 +
a)*sqrt(-(b*c - a*d)/c)/((b^2*c - a*b*d)*x^3 + (a*b*c - a^2*d)*x)) - (2*b*c - 3*a*d)*sqrt(b)*log(-2*b*x^2 - 2
*sqrt(b*x^2 + a)*sqrt(b)*x - a))/d^2, 1/2*(sqrt(b*x^2 + a)*b*d*x + (2*b*c - 3*a*d)*sqrt(-b)*arctan(sqrt(-b)*x/
sqrt(b*x^2 + a)) - (b*c - a*d)*sqrt(-(b*c - a*d)/c)*arctan(1/2*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(
-(b*c - a*d)/c)/((b^2*c - a*b*d)*x^3 + (a*b*c - a^2*d)*x)))/d^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{c + d x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**2)**(3/2)/(c + d*x**2), x)

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{d\,x^2+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/(c + d*x^2),x)

[Out]

int((a + b*x^2)^(3/2)/(c + d*x^2), x)

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