\(\int \frac {(a+b x^2)^{5/2}}{(c+d x^2)^5} \, dx\) [84]

Optimal result

Integrand size = 21, antiderivative size = 249 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c (b c-a d) \left (c+d x^2\right )^4}+\frac {(8 b c-7 a d) x \left (a+b x^2\right )^{5/2}}{48 c^2 (b c-a d) \left (c+d x^2\right )^3}+\frac {5 a (8 b c-7 a d) x \left (a+b x^2\right )^{3/2}}{192 c^3 (b c-a d) \left (c+d x^2\right )^2}+\frac {5 a^2 (8 b c-7 a d) x \sqrt {a+b x^2}}{128 c^4 (b c-a d) \left (c+d x^2\right )}+\frac {5 a^3 (8 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{128 c^{9/2} (b c-a d)^{3/2}} \] Output:

-1/8*d*x*(b*x^2+a)^(7/2)/c/(-a*d+b*c)/(d*x^2+c)^4+1/48*(-7*a*d+8*b*c)*x*(b 
*x^2+a)^(5/2)/c^2/(-a*d+b*c)/(d*x^2+c)^3+5/192*a*(-7*a*d+8*b*c)*x*(b*x^2+a 
)^(3/2)/c^3/(-a*d+b*c)/(d*x^2+c)^2+5/128*a^2*(-7*a*d+8*b*c)*x*(b*x^2+a)^(1 
/2)/c^4/(-a*d+b*c)/(d*x^2+c)+5/128*a^3*(-7*a*d+8*b*c)*arctanh((-a*d+b*c)^( 
1/2)*x/c^(1/2)/(b*x^2+a)^(1/2))/c^(9/2)/(-a*d+b*c)^(3/2)
 

Mathematica [A] (verified)

Time = 10.62 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\frac {x \left (c \left (a+b x^2\right ) \left (-16 b^3 c^3 x^4 \left (4 c+d x^2\right )-8 a b^2 c^2 x^2 \left (26 c^2+11 c d x^2+3 d^2 x^4\right )-2 a^2 b c \left (132 c^3+129 c^2 d x^2+94 c d^2 x^4+25 d^3 x^6\right )+a^3 d \left (279 c^3+511 c^2 d x^2+385 c d^2 x^4+105 d^3 x^6\right )\right )+\frac {15 a^3 (-8 b c+7 a d) \left (c+d x^2\right )^4 \text {arctanh}\left (\sqrt {\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}}\right )}{\sqrt {\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}}}\right )}{384 c^5 (-b c+a d) \sqrt {a+b x^2} \left (c+d x^2\right )^4} \] Input:

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

Output:

(x*(c*(a + b*x^2)*(-16*b^3*c^3*x^4*(4*c + d*x^2) - 8*a*b^2*c^2*x^2*(26*c^2 
 + 11*c*d*x^2 + 3*d^2*x^4) - 2*a^2*b*c*(132*c^3 + 129*c^2*d*x^2 + 94*c*d^2 
*x^4 + 25*d^3*x^6) + a^3*d*(279*c^3 + 511*c^2*d*x^2 + 385*c*d^2*x^4 + 105* 
d^3*x^6)) + (15*a^3*(-8*b*c + 7*a*d)*(c + d*x^2)^4*ArcTanh[Sqrt[((b*c - a* 
d)*x^2)/(c*(a + b*x^2))]])/Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(384* 
c^5*(-(b*c) + a*d)*Sqrt[a + b*x^2]*(c + d*x^2)^4)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {296, 292, 292, 292, 291, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx\)

\(\Big \downarrow \) 296

\(\displaystyle \frac {(8 b c-7 a d) \int \frac {\left (b x^2+a\right )^{5/2}}{\left (d x^2+c\right )^4}dx}{8 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c \left (c+d x^2\right )^4 (b c-a d)}\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {5 a \int \frac {\left (b x^2+a\right )^{3/2}}{\left (d x^2+c\right )^3}dx}{6 c}+\frac {x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}\right )}{8 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c \left (c+d x^2\right )^4 (b c-a d)}\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {5 a \left (\frac {3 a \int \frac {\sqrt {b x^2+a}}{\left (d x^2+c\right )^2}dx}{4 c}+\frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}\right )}{6 c}+\frac {x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}\right )}{8 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c \left (c+d x^2\right )^4 (b c-a d)}\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c}+\frac {x \sqrt {a+b x^2}}{2 c \left (c+d x^2\right )}\right )}{4 c}+\frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}\right )}{6 c}+\frac {x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}\right )}{8 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c \left (c+d x^2\right )^4 (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 c}+\frac {x \sqrt {a+b x^2}}{2 c \left (c+d x^2\right )}\right )}{4 c}+\frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}\right )}{6 c}+\frac {x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}\right )}{8 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c \left (c+d x^2\right )^4 (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} \sqrt {b c-a d}}+\frac {x \sqrt {a+b x^2}}{2 c \left (c+d x^2\right )}\right )}{4 c}+\frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}\right )}{6 c}+\frac {x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}\right )}{8 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^{7/2}}{8 c \left (c+d x^2\right )^4 (b c-a d)}\)

Input:

Int[(a + b*x^2)^(5/2)/(c + d*x^2)^5,x]
 

Output:

-1/8*(d*x*(a + b*x^2)^(7/2))/(c*(b*c - a*d)*(c + d*x^2)^4) + ((8*b*c - 7*a 
*d)*((x*(a + b*x^2)^(5/2))/(6*c*(c + d*x^2)^3) + (5*a*((x*(a + b*x^2)^(3/2 
))/(4*c*(c + d*x^2)^2) + (3*a*((x*Sqrt[a + b*x^2])/(2*c*(c + d*x^2)) + (a* 
ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(2*c^(3/2)*Sqrt[b* 
c - a*d])))/(4*c)))/(6*c)))/(8*c*(b*c - a*d))
 

Defintions of rubi rules used

rule 221

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 291

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

rule 292

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si 
mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( 
a*(p + 1)))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ 
{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt 
Q[q, 0] && NeQ[p, -1]
 

rule 296

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[p, -1]
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(\frac {93 \sqrt {\left (a d -b c \right ) c}\, \left (d \left (\frac {35}{93} d^{3} x^{6}+\frac {385}{279} c \,d^{2} x^{4}+\frac {511}{279} c^{2} d \,x^{2}+c^{3}\right ) a^{3}-\frac {88 c \left (\frac {25}{132} d^{3} x^{6}+\frac {47}{66} c \,d^{2} x^{4}+\frac {43}{44} c^{2} d \,x^{2}+c^{3}\right ) b \,a^{2}}{93}-\frac {208 c^{2} \left (\frac {3}{26} d^{2} x^{4}+\frac {11}{26} c d \,x^{2}+c^{2}\right ) b^{2} x^{2} a}{279}-\frac {64 c^{3} \left (\frac {x^{2} d}{4}+c \right ) b^{3} x^{4}}{279}\right ) x \sqrt {b \,x^{2}+a}-35 a^{3} \left (x^{2} d +c \right )^{4} \left (a d -\frac {8 b c}{7}\right ) \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{128 \sqrt {\left (a d -b c \right ) c}\, \left (a d -b c \right ) c^{4} \left (x^{2} d +c \right )^{4}}\)	\(234\)
default	\(\text {Expression too large to display}\)	\(34027\)

Input:

int((b*x^2+a)^(5/2)/(d*x^2+c)^5,x,method=_RETURNVERBOSE)
 

Output:

1/128*(93*((a*d-b*c)*c)^(1/2)*(d*(35/93*d^3*x^6+385/279*c*d^2*x^4+511/279* 
c^2*d*x^2+c^3)*a^3-88/93*c*(25/132*d^3*x^6+47/66*c*d^2*x^4+43/44*c^2*d*x^2 
+c^3)*b*a^2-208/279*c^2*(3/26*d^2*x^4+11/26*c*d*x^2+c^2)*b^2*x^2*a-64/279* 
c^3*(1/4*x^2*d+c)*b^3*x^4)*x*(b*x^2+a)^(1/2)-35*a^3*(d*x^2+c)^4*(a*d-8/7*b 
*c)*arctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1/2)))/((a*d-b*c)*c)^(1/2)/( 
a*d-b*c)/c^4/(d*x^2+c)^4
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 609 vs. \(2 (221) = 442\).

Time = 0.65 (sec) , antiderivative size = 1258, normalized size of antiderivative = 5.05 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/1536*(15*(8*a^3*b*c^5 - 7*a^4*c^4*d + (8*a^3*b*c*d^4 - 7*a^4*d^5)*x^8 + 
 4*(8*a^3*b*c^2*d^3 - 7*a^4*c*d^4)*x^6 + 6*(8*a^3*b*c^3*d^2 - 7*a^4*c^2*d^ 
3)*x^4 + 4*(8*a^3*b*c^4*d - 7*a^4*c^3*d^2)*x^2)*sqrt(b*c^2 - a*c*d)*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*((2*b*c - a*d)*x^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a))/ 
(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*((16*b^4*c^5*d + 8*a*b^3*c^4*d^2 + 26*a^2 
*b^2*c^3*d^3 - 155*a^3*b*c^2*d^4 + 105*a^4*c*d^5)*x^7 + (64*b^4*c^6 + 24*a 
*b^3*c^5*d + 100*a^2*b^2*c^4*d^2 - 573*a^3*b*c^3*d^3 + 385*a^4*c^2*d^4)*x^ 
5 + (208*a*b^3*c^6 + 50*a^2*b^2*c^5*d - 769*a^3*b*c^4*d^2 + 511*a^4*c^3*d^ 
3)*x^3 + 3*(88*a^2*b^2*c^6 - 181*a^3*b*c^5*d + 93*a^4*c^4*d^2)*x)*sqrt(b*x 
^2 + a))/(b^2*c^11 - 2*a*b*c^10*d + a^2*c^9*d^2 + (b^2*c^7*d^4 - 2*a*b*c^6 
*d^5 + a^2*c^5*d^6)*x^8 + 4*(b^2*c^8*d^3 - 2*a*b*c^7*d^4 + a^2*c^6*d^5)*x^ 
6 + 6*(b^2*c^9*d^2 - 2*a*b*c^8*d^3 + a^2*c^7*d^4)*x^4 + 4*(b^2*c^10*d - 2* 
a*b*c^9*d^2 + a^2*c^8*d^3)*x^2), -1/768*(15*(8*a^3*b*c^5 - 7*a^4*c^4*d + ( 
8*a^3*b*c*d^4 - 7*a^4*d^5)*x^8 + 4*(8*a^3*b*c^2*d^3 - 7*a^4*c*d^4)*x^6 + 6 
*(8*a^3*b*c^3*d^2 - 7*a^4*c^2*d^3)*x^4 + 4*(8*a^3*b*c^4*d - 7*a^4*c^3*d^2) 
*x^2)*sqrt(-b*c^2 + a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)* 
x^2 + a*c)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 + (a*b*c^2 - a^2*c*d)* 
x)) - 2*((16*b^4*c^5*d + 8*a*b^3*c^4*d^2 + 26*a^2*b^2*c^3*d^3 - 155*a^3*b* 
c^2*d^4 + 105*a^4*c*d^5)*x^7 + (64*b^4*c^6 + 24*a*b^3*c^5*d + 100*a^2*b...
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{\left (c + d x^{2}\right )^{5}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{{\left (d x^{2} + c\right )}^{5}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1448 vs. \(2 (221) = 442\).

Time = 0.51 (sec) , antiderivative size = 1448, normalized size of antiderivative = 5.82 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\text {Too large to display} \] Input:

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

Output:

-5/128*(8*a^3*b^(3/2)*c - 7*a^4*sqrt(b)*d)*arctan(1/2*((sqrt(b)*x - sqrt(b 
*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c*d))/((b*c^5 - a*c^4*d) 
*sqrt(-b^2*c^2 + a*b*c*d)) - 1/192*(120*(sqrt(b)*x - sqrt(b*x^2 + a))^14*a 
^3*b^(3/2)*c*d^6 - 105*(sqrt(b)*x - sqrt(b*x^2 + a))^14*a^4*sqrt(b)*d^7 - 
768*(sqrt(b)*x - sqrt(b*x^2 + a))^12*b^(11/2)*c^5*d^2 + 768*(sqrt(b)*x - s 
qrt(b*x^2 + a))^12*a*b^(9/2)*c^4*d^3 + 1680*(sqrt(b)*x - sqrt(b*x^2 + a))^ 
12*a^3*b^(5/2)*c^2*d^5 - 2310*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a^4*b^(3/2) 
*c*d^6 + 735*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a^5*sqrt(b)*d^7 - 2048*(sqrt 
(b)*x - sqrt(b*x^2 + a))^10*b^(13/2)*c^6*d + 2048*(sqrt(b)*x - sqrt(b*x^2 
+ a))^10*a^2*b^(9/2)*c^4*d^3 + 8320*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^3*b 
^(7/2)*c^3*d^4 - 15600*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^4*b^(5/2)*c^2*d^ 
5 + 9800*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^5*b^(3/2)*c*d^6 - 2205*(sqrt(b 
)*x - sqrt(b*x^2 + a))^10*a^6*sqrt(b)*d^7 - 2048*(sqrt(b)*x - sqrt(b*x^2 + 
 a))^8*b^(15/2)*c^7 + 1024*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(13/2)*c^6* 
d - 4864*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^2*b^(11/2)*c^5*d^2 + 21888*(sqr 
t(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(9/2)*c^4*d^3 - 38000*(sqrt(b)*x - sqrt( 
b*x^2 + a))^8*a^4*b^(7/2)*c^3*d^4 + 37400*(sqrt(b)*x - sqrt(b*x^2 + a))^8* 
a^5*b^(5/2)*c^2*d^5 - 18550*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^6*b^(3/2)*c* 
d^6 + 3675*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^7*sqrt(b)*d^7 - 2048*(sqrt(b) 
*x - sqrt(b*x^2 + a))^6*a^2*b^(13/2)*c^6*d - 9472*(sqrt(b)*x - sqrt(b*x...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}}{{\left (d\,x^2+c\right )}^5} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^5} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{\left (d \,x^{2}+c \right )^{5}}d x \] Input:

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

Output:

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