\(\int \frac {c+d x^2}{(e x)^{15/2} (a+b x^2)^{3/4}} \, dx\) [1098]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}+\frac {2 (12 b c-13 a d) \sqrt [4]{a+b x^2}}{13 a^2 e^3 (e x)^{9/2}}-\frac {16 (12 b c-13 a d) \left (a+b x^2\right )^{5/4}}{65 a^3 e^3 (e x)^{9/2}}+\frac {64 (12 b c-13 a d) \left (a+b x^2\right )^{9/4}}{585 a^4 e^3 (e x)^{9/2}} \]

[Out]

-2/13*c*(b*x^2+a)^(1/4)/a/e/(e*x)^(13/2)+2/13*(-13*a*d+12*b*c)*(b*x^2+a)^(1/4)/a^2/e^3/(e*x)^(9/2)-16/65*(-13*
a*d+12*b*c)*(b*x^2+a)^(5/4)/a^3/e^3/(e*x)^(9/2)+64/585*(-13*a*d+12*b*c)*(b*x^2+a)^(9/4)/a^4/e^3/(e*x)^(9/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {464, 279, 270} \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {64 \left (a+b x^2\right )^{9/4} (12 b c-13 a d)}{585 a^4 e^3 (e x)^{9/2}}-\frac {16 \left (a+b x^2\right )^{5/4} (12 b c-13 a d)}{65 a^3 e^3 (e x)^{9/2}}+\frac {2 \sqrt [4]{a+b x^2} (12 b c-13 a d)}{13 a^2 e^3 (e x)^{9/2}}-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}} \]

[In]

Int[(c + d*x^2)/((e*x)^(15/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*c*(a + b*x^2)^(1/4))/(13*a*e*(e*x)^(13/2)) + (2*(12*b*c - 13*a*d)*(a + b*x^2)^(1/4))/(13*a^2*e^3*(e*x)^(9/
2)) - (16*(12*b*c - 13*a*d)*(a + b*x^2)^(5/4))/(65*a^3*e^3*(e*x)^(9/2)) + (64*(12*b*c - 13*a*d)*(a + b*x^2)^(9
/4))/(585*a^4*e^3*(e*x)^(9/2))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}-\frac {(12 b c-13 a d) \int \frac {1}{(e x)^{11/2} \left (a+b x^2\right )^{3/4}} \, dx}{13 a e^2} \\ & = -\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}+\frac {2 (12 b c-13 a d) \sqrt [4]{a+b x^2}}{13 a^2 e^3 (e x)^{9/2}}+\frac {(8 (12 b c-13 a d)) \int \frac {\sqrt [4]{a+b x^2}}{(e x)^{11/2}} \, dx}{13 a^2 e^2} \\ & = -\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}+\frac {2 (12 b c-13 a d) \sqrt [4]{a+b x^2}}{13 a^2 e^3 (e x)^{9/2}}-\frac {16 (12 b c-13 a d) \left (a+b x^2\right )^{5/4}}{65 a^3 e^3 (e x)^{9/2}}-\frac {(32 (12 b c-13 a d)) \int \frac {\left (a+b x^2\right )^{5/4}}{(e x)^{11/2}} \, dx}{65 a^3 e^2} \\ & = -\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}+\frac {2 (12 b c-13 a d) \sqrt [4]{a+b x^2}}{13 a^2 e^3 (e x)^{9/2}}-\frac {16 (12 b c-13 a d) \left (a+b x^2\right )^{5/4}}{65 a^3 e^3 (e x)^{9/2}}+\frac {64 (12 b c-13 a d) \left (a+b x^2\right )^{9/4}}{585 a^4 e^3 (e x)^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 x \sqrt [4]{a+b x^2} \left (45 a^3 c-60 a^2 b c x^2+65 a^3 d x^2+96 a b^2 c x^4-104 a^2 b d x^4-384 b^3 c x^6+416 a b^2 d x^6\right )}{585 a^4 (e x)^{15/2}} \]

[In]

Integrate[(c + d*x^2)/((e*x)^(15/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*x*(a + b*x^2)^(1/4)*(45*a^3*c - 60*a^2*b*c*x^2 + 65*a^3*d*x^2 + 96*a*b^2*c*x^4 - 104*a^2*b*d*x^4 - 384*b^3
*c*x^6 + 416*a*b^2*d*x^6))/(585*a^4*(e*x)^(15/2))

Maple [A] (verified)

Time = 3.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61

method	result	size
gosper	\(-\frac {2 x \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (416 a \,b^{2} d \,x^{6}-384 b^{3} c \,x^{6}-104 a^{2} b d \,x^{4}+96 a \,b^{2} c \,x^{4}+65 a^{3} d \,x^{2}-60 a^{2} b c \,x^{2}+45 c \,a^{3}\right )}{585 a^{4} \left (e x \right )^{\frac {15}{2}}}\)	\(86\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (416 a \,b^{2} d \,x^{6}-384 b^{3} c \,x^{6}-104 a^{2} b d \,x^{4}+96 a \,b^{2} c \,x^{4}+65 a^{3} d \,x^{2}-60 a^{2} b c \,x^{2}+45 c \,a^{3}\right )}{585 e^{7} \sqrt {e x}\, a^{4} x^{6}}\)	\(91\)

[In]

int((d*x^2+c)/(e*x)^(15/2)/(b*x^2+a)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-2/585*x*(b*x^2+a)^(1/4)*(416*a*b^2*d*x^6-384*b^3*c*x^6-104*a^2*b*d*x^4+96*a*b^2*c*x^4+65*a^3*d*x^2-60*a^2*b*c
*x^2+45*a^3*c)/a^4/(e*x)^(15/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \, {\left (32 \, {\left (12 \, b^{3} c - 13 \, a b^{2} d\right )} x^{6} - 8 \, {\left (12 \, a b^{2} c - 13 \, a^{2} b d\right )} x^{4} - 45 \, a^{3} c + 5 \, {\left (12 \, a^{2} b c - 13 \, a^{3} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{585 \, a^{4} e^{8} x^{7}} \]

[In]

integrate((d*x^2+c)/(e*x)^(15/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

2/585*(32*(12*b^3*c - 13*a*b^2*d)*x^6 - 8*(12*a*b^2*c - 13*a^2*b*d)*x^4 - 45*a^3*c + 5*(12*a^2*b*c - 13*a^3*d)
*x^2)*(b*x^2 + a)^(1/4)*sqrt(e*x)/(a^4*e^8*x^7)

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\text {Timed out} \]

[In]

integrate((d*x**2+c)/(e*x)**(15/2)/(b*x**2+a)**(3/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {15}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(15/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(15/2)), x)

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {15}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(15/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(15/2)), x)

Mupad [B] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,c}{13\,a\,e^7}+\frac {x^2\,\left (130\,a^3\,d-120\,a^2\,b\,c\right )}{585\,a^4\,e^7}-\frac {x^6\,\left (768\,b^3\,c-832\,a\,b^2\,d\right )}{585\,a^4\,e^7}-\frac {16\,b\,x^4\,\left (13\,a\,d-12\,b\,c\right )}{585\,a^3\,e^7}\right )}{x^6\,\sqrt {e\,x}} \]

[In]

int((c + d*x^2)/((e*x)^(15/2)*(a + b*x^2)^(3/4)),x)

[Out]

-((a + b*x^2)^(1/4)*((2*c)/(13*a*e^7) + (x^2*(130*a^3*d - 120*a^2*b*c))/(585*a^4*e^7) - (x^6*(768*b^3*c - 832*
a*b^2*d))/(585*a^4*e^7) - (16*b*x^4*(13*a*d - 12*b*c))/(585*a^3*e^7)))/(x^6*(e*x)^(1/2))