\(\int \frac {(c+d x^2)^{3/2}}{a+b x^2} \, dx\) [689]

Optimal result

Integrand size = 21, antiderivative size = 113 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {d x \sqrt {c+d x^2}}{2 b}+\frac {(b c-a d)^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} b^2}+\frac {\sqrt {d} (3 b c-2 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {427, 537, 223, 212, 385, 211} \[ \int \frac {\left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {(b c-a d)^{3/2} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} b^2}+\frac {\sqrt {d} (3 b c-2 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2}+\frac {d x \sqrt {c+d x^2}}{2 b} \]

[In]

Int[(c + d*x^2)^(3/2)/(a + b*x^2),x]

[Out]

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

Rule 211

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

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {b d x \sqrt {c+d x^2}-\frac {2 (b c-a d)^{3/2} \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{\sqrt {a}}+\sqrt {d} (-3 b c+2 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.00 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(-\frac {-\left (a d -b c \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )+\sqrt {\left (a d -b c \right ) a}\, \left (\left (d^{\frac {3}{2}} a -\frac {3 b \sqrt {d}\, c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )-\frac {\sqrt {d \,x^{2}+c}\, b d x}{2}\right )}{\sqrt {\left (a d -b c \right ) a}\, b^{2}}\)	\(117\)
risch	\(\frac {d x \sqrt {d \,x^{2}+c}}{2 b}-\frac {\frac {\sqrt {d}\, \left (2 a d -3 b c \right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{b}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{\sqrt {-a b}\, b \sqrt {-\frac {a d -b c}{b}}}-\frac {\left (-a^{2} d^{2}+2 a b c d -b^{2} c^{2}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{\sqrt {-a b}\, b \sqrt {-\frac {a d -b c}{b}}}}{2 b}\)	\(410\)
default	\(\text {Expression too large to display}\)	\(1243\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 721, normalized size of antiderivative = 6.38 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\left [\frac {2 \, \sqrt {d x^{2} + c} b d x - {\left (3 \, b c - 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (b c - a d\right )} \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 \, b^{2}}, \frac {2 \, \sqrt {d x^{2} + c} b d x - 2 \, {\left (3 \, b c - 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (b c - a d\right )} \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 \, b^{2}}, \frac {2 \, \sqrt {d x^{2} + c} b d x + 2 \, {\left (b c - a d\right )} \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 ) - {\left (3 \, b c - 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, b^{2}}, \frac {\sqrt {d x^{2} + c} b d x - {\left (3 \, b c - 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b c - a d\right )} \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 \, b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\int \frac {{\left (d\,x^2+c\right )}^{3/2}}{b\,x^2+a} \,d x \]

[In]

int((c + d*x^2)^(3/2)/(a + b*x^2),x)

[Out]

int((c + d*x^2)^(3/2)/(a + b*x^2), x)