3.69 \(\int \frac{(a+b x^2)^{5/2}}{(c+d x^2)^4} \, dx\)

Optimal. Leaf size=144 \[ \frac{5 a^2 x \sqrt{a+b x^2}}{16 c^3 \left (c+d x^2\right )}+\frac{5 a^3 \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{16 c^{7/2} \sqrt{b c-a d}}+\frac{5 a x \left (a+b x^2\right )^{3/2}}{24 c^2 \left (c+d x^2\right )^2}+\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3} \]

[Out]

(x*(a + b*x^2)^(5/2))/(6*c*(c + d*x^2)^3) + (5*a*x*(a + b*x^2)^(3/2))/(24*c^2*(c + d*x^2)^2) + (5*a^2*x*Sqrt[a
+ b*x^2])/(16*c^3*(c + d*x^2)) + (5*a^3*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(16*c^(7/2)*S
qrt[b*c - a*d])

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Rubi [A]  time = 0.0732243, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {378, 377, 208} \[ \frac{5 a^2 x \sqrt{a+b x^2}}{16 c^3 \left (c+d x^2\right )}+\frac{5 a^3 \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{16 c^{7/2} \sqrt{b c-a d}}+\frac{5 a x \left (a+b x^2\right )^{3/2}}{24 c^2 \left (c+d x^2\right )^2}+\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/(c + d*x^2)^4,x]

[Out]

(x*(a + b*x^2)^(5/2))/(6*c*(c + d*x^2)^3) + (5*a*x*(a + b*x^2)^(3/2))/(24*c^2*(c + d*x^2)^2) + (5*a^2*x*Sqrt[a
+ b*x^2])/(16*c^3*(c + d*x^2)) + (5*a^3*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(16*c^(7/2)*S
qrt[b*c - a*d])

Rule 378

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

Rule 377

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 208

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{5/2}}{\left (c+d x^2\right )^4} \, dx &=\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}+\frac{(5 a) \int \frac{\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx}{6 c}\\ &=\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}+\frac{5 a x \left (a+b x^2\right )^{3/2}}{24 c^2 \left (c+d x^2\right )^2}+\frac{\left (5 a^2\right ) \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^2} \, dx}{8 c^2}\\ &=\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}+\frac{5 a x \left (a+b x^2\right )^{3/2}}{24 c^2 \left (c+d x^2\right )^2}+\frac{5 a^2 x \sqrt{a+b x^2}}{16 c^3 \left (c+d x^2\right )}+\frac{\left (5 a^3\right ) \int \frac{1}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{16 c^3}\\ &=\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}+\frac{5 a x \left (a+b x^2\right )^{3/2}}{24 c^2 \left (c+d x^2\right )^2}+\frac{5 a^2 x \sqrt{a+b x^2}}{16 c^3 \left (c+d x^2\right )}+\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 c^3}\\ &=\frac{x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3}+\frac{5 a x \left (a+b x^2\right )^{3/2}}{24 c^2 \left (c+d x^2\right )^2}+\frac{5 a^2 x \sqrt{a+b x^2}}{16 c^3 \left (c+d x^2\right )}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{c} \sqrt{a+b x^2}}\right )}{16 c^{7/2} \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.777666, size = 201, normalized size = 1.4 \[ \frac{x \sqrt{a+b x^2} \left (\frac{\sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \left (a^2 \left (33 c^2+40 c d x^2+15 d^2 x^4\right )+2 a b c x^2 \left (13 c+5 d x^2\right )+8 b^2 c^2 x^4\right )}{\left (c+d x^2\right )^2 \sqrt{\frac{d x^2}{c}+1}}+\frac{15 a^2 \sin ^{-1}\left (\frac{\sqrt{x^2 \left (\frac{d}{c}-\frac{b}{a}\right )}}{\sqrt{\frac{d x^2}{c}+1}}\right )}{\sqrt{\frac{x^2 (a d-b c)}{a c}}}\right )}{48 c^4 \sqrt{\frac{b x^2}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^2)^(5/2)/(c + d*x^2)^4,x]

[Out]

(x*Sqrt[a + b*x^2]*((Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*(8*b^2*c^2*x^4 + 2*a*b*c*x^2*(13*c + 5*d*x^2) + a^2
*(33*c^2 + 40*c*d*x^2 + 15*d^2*x^4)))/((c + d*x^2)^2*Sqrt[1 + (d*x^2)/c]) + (15*a^2*ArcSin[Sqrt[(-(b/a) + d/c)
*x^2]/Sqrt[1 + (d*x^2)/c]])/Sqrt[((-(b*c) + a*d)*x^2)/(a*c)]))/(48*c^4*Sqrt[1 + (b*x^2)/a])

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Maple [B]  time = 0.033, size = 21220, normalized size = 147.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/(d*x^2+c)^4,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{5}{2}}}{{\left (d x^{2} + c\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/(d*x^2+c)^4,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(5/2)/(d*x^2 + c)^4, x)

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Fricas [B]  time = 3.20487, size = 1438, normalized size = 9.99 \begin{align*} \left [\frac{15 \,{\left (a^{3} d^{3} x^{6} + 3 \, a^{3} c d^{2} x^{4} + 3 \, a^{3} c^{2} d x^{2} + a^{3} c^{3}\right )} \sqrt{b c^{2} - a c d} \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 ({\left (2 \, b c - a d\right )} x^{3} + a c x\right )} \sqrt{b c^{2} - a c d} \sqrt{b x^{2} + a}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right ) + 4 \,{\left ({\left (8 \, b^{3} c^{4} + 2 \, a b^{2} c^{3} d + 5 \, a^{2} b c^{2} d^{2} - 15 \, a^{3} c d^{3}\right )} x^{5} + 2 \,{\left (13 \, a b^{2} c^{4} + 7 \, a^{2} b c^{3} d - 20 \, a^{3} c^{2} d^{2}\right )} x^{3} + 33 \,{\left (a^{2} b c^{4} - a^{3} c^{3} d\right )} x\right )} \sqrt{b x^{2} + a}}{192 \,{\left (b c^{8} - a c^{7} d +{\left (b c^{5} d^{3} - a c^{4} d^{4}\right )} x^{6} + 3 \,{\left (b c^{6} d^{2} - a c^{5} d^{3}\right )} x^{4} + 3 \,{\left (b c^{7} d - a c^{6} d^{2}\right )} x^{2}\right )}}, -\frac{15 \,{\left (a^{3} d^{3} x^{6} + 3 \, a^{3} c d^{2} x^{4} + 3 \, a^{3} c^{2} d x^{2} + a^{3} c^{3}\right )} \sqrt{-b c^{2} + a c d} \arctan \left (\frac{\sqrt{-b c^{2} + a c d}{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt{b x^{2} + a}}{2 \,{\left ({\left (b^{2} c^{2} - a b c d\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \,{\left ({\left (8 \, b^{3} c^{4} + 2 \, a b^{2} c^{3} d + 5 \, a^{2} b c^{2} d^{2} - 15 \, a^{3} c d^{3}\right )} x^{5} + 2 \,{\left (13 \, a b^{2} c^{4} + 7 \, a^{2} b c^{3} d - 20 \, a^{3} c^{2} d^{2}\right )} x^{3} + 33 \,{\left (a^{2} b c^{4} - a^{3} c^{3} d\right )} x\right )} \sqrt{b x^{2} + a}}{96 \,{\left (b c^{8} - a c^{7} d +{\left (b c^{5} d^{3} - a c^{4} d^{4}\right )} x^{6} + 3 \,{\left (b c^{6} d^{2} - a c^{5} d^{3}\right )} x^{4} + 3 \,{\left (b c^{7} d - a c^{6} d^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/(d*x^2+c)^4,x, algorithm="fricas")

[Out]

[1/192*(15*(a^3*d^3*x^6 + 3*a^3*c*d^2*x^4 + 3*a^3*c^2*d*x^2 + a^3*c^3)*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*((8*b^3*c^4 + 2*a*b^2*c^3*d + 5*a^2*b*c^2*d^2 - 15*a
^3*c*d^3)*x^5 + 2*(13*a*b^2*c^4 + 7*a^2*b*c^3*d - 20*a^3*c^2*d^2)*x^3 + 33*(a^2*b*c^4 - a^3*c^3*d)*x)*sqrt(b*x
^2 + a))/(b*c^8 - a*c^7*d + (b*c^5*d^3 - a*c^4*d^4)*x^6 + 3*(b*c^6*d^2 - a*c^5*d^3)*x^4 + 3*(b*c^7*d - a*c^6*d
^2)*x^2), -1/96*(15*(a^3*d^3*x^6 + 3*a^3*c*d^2*x^4 + 3*a^3*c^2*d*x^2 + a^3*c^3)*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*((8*b^3*c^4 + 2*a*b^2*c^3*d + 5*a^2*b*c^2*d^2 - 15*a^3*c*d^3)*x^5 + 2*(13*a*b^2*c^4 + 7*a^2*b*c^3*d
- 20*a^3*c^2*d^2)*x^3 + 33*(a^2*b*c^4 - a^3*c^3*d)*x)*sqrt(b*x^2 + a))/(b*c^8 - a*c^7*d + (b*c^5*d^3 - a*c^4*d
^4)*x^6 + 3*(b*c^6*d^2 - a*c^5*d^3)*x^4 + 3*(b*c^7*d - a*c^6*d^2)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/(d*x**2+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 22.0103, size = 1142, normalized size = 7.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/(d*x^2+c)^4,x, algorithm="giac")

[Out]

-5/16*a^3*sqrt(b)*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))/(sqrt
(-b^2*c^2 + a*b*c*d)*c^3) + 1/24*(48*(sqrt(b)*x - sqrt(b*x^2 + a))^10*b^(7/2)*c^3*d^2 - 15*(sqrt(b)*x - sqrt(b
*x^2 + a))^10*a^3*sqrt(b)*d^5 + 192*(sqrt(b)*x - sqrt(b*x^2 + a))^8*b^(9/2)*c^4*d + 48*(sqrt(b)*x - sqrt(b*x^2
+ a))^8*a*b^(7/2)*c^3*d^2 - 150*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(3/2)*c*d^4 + 75*(sqrt(b)*x - sqrt(b*x^
2 + a))^8*a^4*sqrt(b)*d^5 + 256*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(11/2)*c^5 - 64*(sqrt(b)*x - sqrt(b*x^2 + a)
)^6*a*b^(9/2)*c^4*d + 288*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^2*b^(7/2)*c^3*d^2 - 440*(sqrt(b)*x - sqrt(b*x^2 +
a))^6*a^3*b^(5/2)*c^2*d^3 + 440*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^4*b^(3/2)*c*d^4 - 150*(sqrt(b)*x - sqrt(b*x^
2 + a))^6*a^5*sqrt(b)*d^5 + 192*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(9/2)*c^4*d + 48*(sqrt(b)*x - sqrt(b*x^2
+ a))^4*a^3*b^(7/2)*c^3*d^2 + 360*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^4*b^(5/2)*c^2*d^3 - 420*(sqrt(b)*x - sqrt
(b*x^2 + a))^4*a^5*b^(3/2)*c*d^4 + 150*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^6*sqrt(b)*d^5 + 48*(sqrt(b)*x - sqrt(
b*x^2 + a))^2*a^4*b^(7/2)*c^3*d^2 + 72*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^5*b^(5/2)*c^2*d^3 + 120*(sqrt(b)*x -
sqrt(b*x^2 + a))^2*a^6*b^(3/2)*c*d^4 - 75*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^7*sqrt(b)*d^5 + 8*a^6*b^(5/2)*c^2*
d^3 + 10*a^7*b^(3/2)*c*d^4 + 15*a^8*sqrt(b)*d^5)/(((sqrt(b)*x - sqrt(b*x^2 + a))^4*d + 4*(sqrt(b)*x - sqrt(b*x
^2 + a))^2*b*c - 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*d + a^2*d)^3*c^3*d^3)