\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{7/2}} \, dx\) [86]

Optimal result

Integrand size = 31, antiderivative size = 216 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=-\frac {8 c^2 (B c-A d) \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}-\frac {4 c (B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2 (c+d x)^{3/2}}-\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2 (c+d x)^{5/2}}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{7 d^2 (c+d x)^{7/2}}+\frac {8 \sqrt {2} c^{5/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:

-8*c^2*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)-4/3*c*(-A*d+B*c)* 
(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(3/2)-2/5*(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2) 
/d^2/(d*x+c)^(5/2)-2/7*B*(-d^2*x^2+c^2)^(7/2)/d^2/(d*x+c)^(7/2)+8*2^(1/2)* 
c^(5/2)*(-A*d+B*c)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1 
/2))/d^2
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\frac {2 \left (\frac {\sqrt {c^2-d^2 x^2} \left (7 A d \left (73 c^2-16 c d x+3 d^2 x^2\right )+B \left (-526 c^3+157 c^2 d x-66 c d^2 x^2+15 d^3 x^3\right )\right )}{\sqrt {c+d x}}+420 \sqrt {2} c^{5/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{105 d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(7/2),x]
 

Output:

(2*((Sqrt[c^2 - d^2*x^2]*(7*A*d*(73*c^2 - 16*c*d*x + 3*d^2*x^2) + B*(-526* 
c^3 + 157*c^2*d*x - 66*c*d^2*x^2 + 15*d^3*x^3)))/Sqrt[c + d*x] + 420*Sqrt[ 
2]*c^(5/2)*(B*c - A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - 
d^2*x^2]]))/(105*d^2)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {672, 466, 466, 466, 471, 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.

\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx\)

\(\Big \downarrow \) 672

\(\displaystyle -\frac {(B c-A d) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}}dx}{d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{7 d^2 (c+d x)^{7/2}}\)

\(\Big \downarrow \) 466

\(\displaystyle -\frac {(B c-A d) \left (2 c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}}dx+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{5/2}}\right )}{d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{7 d^2 (c+d x)^{7/2}}\)

\(\Big \downarrow \) 466

\(\displaystyle -\frac {(B c-A d) \left (2 c \left (2 c \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{5/2}}\right )}{d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{7 d^2 (c+d x)^{7/2}}\)

\(\Big \downarrow \) 466

\(\displaystyle -\frac {(B c-A d) \left (2 c \left (2 c \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{5/2}}\right )}{d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{7 d^2 (c+d x)^{7/2}}\)

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(B c-A d) \left (2 c \left (2 c \left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{5/2}}\right )}{d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{7 d^2 (c+d x)^{7/2}}\)

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(7/2),x]
 

Output:

(-2*B*(c^2 - d^2*x^2)^(7/2))/(7*d^2*(c + d*x)^(7/2)) - ((B*c - A*d)*((2*(c 
^2 - d^2*x^2)^(5/2))/(5*d*(c + d*x)^(5/2)) + 2*c*((2*(c^2 - d^2*x^2)^(3/2) 
)/(3*d*(c + d*x)^(3/2)) + 2*c*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - 
 (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + 
d*x])])/d))))/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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
 

rule 471

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

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^ 
2 + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {2 \left (15 B \,d^{3} x^{3}+21 A \,d^{3} x^{2}-66 B c \,d^{2} x^{2}-112 A c \,d^{2} x +157 B \,c^{2} d x +511 A \,c^{2} d -526 B \,c^{3}\right ) \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{105 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {8 c^{\frac {5}{2}} \left (A d -B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(195\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-15 B \,d^{3} x^{3} \sqrt {-d x +c}+420 A \,c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d -21 A \,d^{3} x^{2} \sqrt {-d x +c}-420 B \,c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )+66 B c \,d^{2} x^{2} \sqrt {-d x +c}+112 A c \,d^{2} x \sqrt {-d x +c}-157 B \,c^{2} d x \sqrt {-d x +c}-511 A \,c^{2} d \sqrt {-d x +c}+526 B \,c^{3} \sqrt {-d x +c}\right )}{105 \sqrt {d x +c}\, \sqrt {-d x +c}\, d^{2}}\)	\(202\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/105/d^2*(15*B*d^3*x^3+21*A*d^3*x^2-66*B*c*d^2*x^2-112*A*c*d^2*x+157*B*c^ 
2*d*x+511*A*c^2*d-526*B*c^3)*(-d*x+c)^(1/2)*((-d^2*x^2+c^2)/(d*x+c))^(1/2) 
/(-d^2*x^2+c^2)^(1/2)*(d*x+c)^(1/2)-8*c^(5/2)*(A*d-B*c)/d^2*2^(1/2)*arctan 
h(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*((-d^2*x^2+c^2)/(d*x+c))^(1/2)/(-d^2 
*x^2+c^2)^(1/2)*(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.79 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\left [-\frac {2 \, {\left (210 \, \sqrt {2} {\left (B c^{4} - A c^{3} d + {\left (B c^{3} d - A c^{2} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - {\left (15 \, B d^{3} x^{3} - 526 \, B c^{3} + 511 \, A c^{2} d - 3 \, {\left (22 \, B c d^{2} - 7 \, A d^{3}\right )} x^{2} + {\left (157 \, B c^{2} d - 112 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{105 \, {\left (d^{3} x + c d^{2}\right )}}, -\frac {2 \, {\left (420 \, \sqrt {2} {\left (B c^{4} - A c^{3} d + {\left (B c^{3} d - A c^{2} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - {\left (15 \, B d^{3} x^{3} - 526 \, B c^{3} + 511 \, A c^{2} d - 3 \, {\left (22 \, B c d^{2} - 7 \, A d^{3}\right )} x^{2} + {\left (157 \, B c^{2} d - 112 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{105 \, {\left (d^{3} x + c d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(7/2),x, algorithm="fricas" 
)
 

Output:

[-2/105*(210*sqrt(2)*(B*c^4 - A*c^3*d + (B*c^3*d - A*c^2*d^2)*x)*sqrt(c)*l 
og(-(d^2*x^2 - 2*c*d*x + 2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt 
(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) - (15*B*d^3*x^3 - 526*B*c^3 + 511* 
A*c^2*d - 3*(22*B*c*d^2 - 7*A*d^3)*x^2 + (157*B*c^2*d - 112*A*c*d^2)*x)*sq 
rt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^3*x + c*d^2), -2/105*(420*sqrt(2)*(B* 
c^4 - A*c^3*d + (B*c^3*d - A*c^2*d^2)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt( 
-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) - (15*B*d^3*x^3 - 52 
6*B*c^3 + 511*A*c^2*d - 3*(22*B*c*d^2 - 7*A*d^3)*x^2 + (157*B*c^2*d - 112* 
A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^3*x + c*d^2)]
 

Sympy [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(7/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x)/(c + d*x)**(7/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(7/2),x, algorithm="maxima" 
)
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(7/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.77 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\frac {2 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B - 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c - 70 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{2} - 420 \, \sqrt {-d x + c} B c^{3} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A d + 70 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c d + 420 \, \sqrt {-d x + c} A c^{2} d - \frac {420 \, \sqrt {2} {\left (B c^{4} - A c^{3} d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}}\right )}}{105 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

2/105*(15*(d*x - c)^3*sqrt(-d*x + c)*B - 21*(d*x - c)^2*sqrt(-d*x + c)*B*c 
 - 70*(-d*x + c)^(3/2)*B*c^2 - 420*sqrt(-d*x + c)*B*c^3 + 21*(d*x - c)^2*s 
qrt(-d*x + c)*A*d + 70*(-d*x + c)^(3/2)*A*c*d + 420*sqrt(-d*x + c)*A*c^2*d 
 - 420*sqrt(2)*(B*c^4 - A*c^3*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c 
))/sqrt(-c))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{7/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(7/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{7/2}} \, dx=\frac {\frac {146 \sqrt {-d x +c}\, a \,c^{2} d}{15}-\frac {32 \sqrt {-d x +c}\, a c \,d^{2} x}{15}+\frac {2 \sqrt {-d x +c}\, a \,d^{3} x^{2}}{5}-\frac {1052 \sqrt {-d x +c}\, b \,c^{3}}{105}+\frac {314 \sqrt {-d x +c}\, b \,c^{2} d x}{105}-\frac {44 \sqrt {-d x +c}\, b c \,d^{2} x^{2}}{35}+\frac {2 \sqrt {-d x +c}\, b \,d^{3} x^{3}}{7}+8 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,c^{2} d -8 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{3}-\frac {184 \sqrt {c}\, \sqrt {2}\, a \,c^{2} d}{15}+\frac {1528 \sqrt {c}\, \sqrt {2}\, b \,c^{3}}{105}}{d^{2}} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(7/2),x)
 

Output:

(2*(511*sqrt(c - d*x)*a*c**2*d - 112*sqrt(c - d*x)*a*c*d**2*x + 21*sqrt(c 
- d*x)*a*d**3*x**2 - 526*sqrt(c - d*x)*b*c**3 + 157*sqrt(c - d*x)*b*c**2*d 
*x - 66*sqrt(c - d*x)*b*c*d**2*x**2 + 15*sqrt(c - d*x)*b*d**3*x**3 + 420*s 
qrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d 
- 420*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b* 
c**3 - 644*sqrt(c)*sqrt(2)*a*c**2*d + 764*sqrt(c)*sqrt(2)*b*c**3))/(105*d* 
*2)