3.86 \(\int \frac{1}{(a+b x^2)^{3/2} (c+d x^2)} \, dx\)

Optimal. Leaf size=79 \[ \frac{b x}{a \sqrt{a+b x^2} (b c-a d)}-\frac{d \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} (b c-a d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0386197, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {382, 377, 208} \[ \frac{b x}{a \sqrt{a+b x^2} (b c-a d)}-\frac{d \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} (b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(3/2)*(c + d*x^2)),x]

[Out]

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

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && 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{1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )} \, dx &=\frac{b x}{a (b c-a d) \sqrt{a+b x^2}}-\frac{d \int \frac{1}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{b c-a d}\\ &=\frac{b x}{a (b c-a d) \sqrt{a+b x^2}}-\frac{d \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{b c-a d}\\ &=\frac{b x}{a (b c-a d) \sqrt{a+b x^2}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.682355, size = 309, normalized size = 3.91 \[ \frac{x \left (2 d x^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+2 c \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-10 d x^2 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}-15 c \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}+10 d x^2 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )+15 c \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )\right )}{5 c^2 \left (a+b x^2\right )^{3/2} \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)^(3/2)*(c + d*x^2)),x]

[Out]

(x*(-15*c*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 10*d*x^2*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 15*c*Ar
cTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + 10*d*x^2*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + 2
*c*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]
+ 2*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, ((b*c - a*d)*x^2)/(c*(a +
b*x^2))]))/(5*c^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2)*(a + b*x^2)^(3/2))

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Maple [B]  time = 0.013, size = 618, normalized size = 7.8 \begin{align*}{\frac{d}{2\,ad-2\,bc}{\frac{1}{\sqrt{-cd}}}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{d}\sqrt{-cd}} \right ) ^{2}b+2\,{\frac{b\sqrt{-cd}}{d} \left ( x-{\frac{\sqrt{-cd}}{d}} \right ) }+{\frac{ad-bc}{d}}}}}}-{\frac{bx}{ \left ( 2\,ad-2\,bc \right ) a}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{d}\sqrt{-cd}} \right ) ^{2}b+2\,{\frac{b\sqrt{-cd}}{d} \left ( x-{\frac{\sqrt{-cd}}{d}} \right ) }+{\frac{ad-bc}{d}}}}}}-{\frac{d}{2\,ad-2\,bc}\ln \left ({ \left ( 2\,{\frac{ad-bc}{d}}+2\,{\frac{b\sqrt{-cd}}{d} \left ( x-{\frac{\sqrt{-cd}}{d}} \right ) }+2\,\sqrt{{\frac{ad-bc}{d}}}\sqrt{ \left ( x-{\frac{\sqrt{-cd}}{d}} \right ) ^{2}b+2\,{\frac{b\sqrt{-cd}}{d} \left ( x-{\frac{\sqrt{-cd}}{d}} \right ) }+{\frac{ad-bc}{d}}} \right ) \left ( x-{\frac{1}{d}\sqrt{-cd}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-cd}}}{\frac{1}{\sqrt{{\frac{ad-bc}{d}}}}}}-{\frac{d}{2\,ad-2\,bc}{\frac{1}{\sqrt{-cd}}}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{d}\sqrt{-cd}} \right ) ^{2}b-2\,{\frac{b\sqrt{-cd}}{d} \left ( x+{\frac{\sqrt{-cd}}{d}} \right ) }+{\frac{ad-bc}{d}}}}}}-{\frac{bx}{ \left ( 2\,ad-2\,bc \right ) a}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{d}\sqrt{-cd}} \right ) ^{2}b-2\,{\frac{b\sqrt{-cd}}{d} \left ( x+{\frac{\sqrt{-cd}}{d}} \right ) }+{\frac{ad-bc}{d}}}}}}+{\frac{d}{2\,ad-2\,bc}\ln \left ({ \left ( 2\,{\frac{ad-bc}{d}}-2\,{\frac{b\sqrt{-cd}}{d} \left ( x+{\frac{\sqrt{-cd}}{d}} \right ) }+2\,\sqrt{{\frac{ad-bc}{d}}}\sqrt{ \left ( x+{\frac{\sqrt{-cd}}{d}} \right ) ^{2}b-2\,{\frac{b\sqrt{-cd}}{d} \left ( x+{\frac{\sqrt{-cd}}{d}} \right ) }+{\frac{ad-bc}{d}}} \right ) \left ( x+{\frac{1}{d}\sqrt{-cd}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-cd}}}{\frac{1}{\sqrt{{\frac{ad-bc}{d}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(3/2)/(d*x^2+c),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.36161, size = 911, normalized size = 11.53 \begin{align*} \left [\frac{4 \,{\left (b^{2} c^{2} - a b c d\right )} \sqrt{b x^{2} + a} x -{\left (a b d x^{2} + a^{2} d\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 (a^{2} b^{2} c^{3} - 2 \, a^{3} b c^{2} d + a^{4} c d^{2} +{\left (a b^{3} c^{3} - 2 \, a^{2} b^{2} c^{2} d + a^{3} b c d^{2}\right )} x^{2}\right )}}, \frac{2 \,{\left (b^{2} c^{2} - a b c d\right )} \sqrt{b x^{2} + a} x +{\left (a b d x^{2} + a^{2} d\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 (a^{2} b^{2} c^{3} - 2 \, a^{3} b c^{2} d + a^{4} c d^{2} +{\left (a b^{3} c^{3} - 2 \, a^{2} b^{2} c^{2} d + a^{3} b c d^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{2}\right )^{\frac{3}{2}} \left (c + d x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(3/2)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(3/2)*(c + d*x**2)), x)

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Giac [A]  time = 1.14144, size = 144, normalized size = 1.82 \begin{align*} -\frac{\sqrt{b} d \arctan \left (-\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt{-b^{2} c^{2} + a b c d}}\right )}{\sqrt{-b^{2} c^{2} + a b c d}{\left (b c - a d\right )}} + \frac{b x}{{\left (a b c - a^{2} d\right )} \sqrt{b x^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c),x, algorithm="giac")

[Out]

-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))/(sqrt(-b^2*
c^2 + a*b*c*d)*(b*c - a*d)) + b*x/((a*b*c - a^2*d)*sqrt(b*x^2 + a))