3.281 \(\int e^{-\tan ^{-1}(a x)} (c+a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=98 \[ -\frac{\left (\frac{1}{13}-\frac{5 i}{13}\right ) 2^{\frac{3}{2}+\frac{i}{2}} c (1-i a x)^{\frac{5}{2}-\frac{i}{2}} \sqrt{a^2 c x^2+c} \, _2F_1\left (-\frac{3}{2}-\frac{i}{2},\frac{5}{2}-\frac{i}{2};\frac{7}{2}-\frac{i}{2};\frac{1}{2} (1-i a x)\right )}{a \sqrt{a^2 x^2+1}} \]

[Out]

((-1/13 + (5*I)/13)*2^(3/2 + I/2)*c*(1 - I*a*x)^(5/2 - I/2)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[-3/2 - I/2,
5/2 - I/2, 7/2 - I/2, (1 - I*a*x)/2])/(a*Sqrt[1 + a^2*x^2])

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Rubi [A]  time = 0.0764833, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {5076, 5073, 69} \[ -\frac{\left (\frac{1}{13}-\frac{5 i}{13}\right ) 2^{\frac{3}{2}+\frac{i}{2}} c (1-i a x)^{\frac{5}{2}-\frac{i}{2}} \sqrt{a^2 c x^2+c} \, _2F_1\left (-\frac{3}{2}-\frac{i}{2},\frac{5}{2}-\frac{i}{2};\frac{7}{2}-\frac{i}{2};\frac{1}{2} (1-i a x)\right )}{a \sqrt{a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^(3/2)/E^ArcTan[a*x],x]

[Out]

((-1/13 + (5*I)/13)*2^(3/2 + I/2)*c*(1 - I*a*x)^(5/2 - I/2)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[-3/2 - I/2,
5/2 - I/2, 7/2 - I/2, (1 - I*a*x)/2])/(a*Sqrt[1 + a^2*x^2])

Rule 5076

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d*x^2)^FracP
art[p])/(1 + a^2*x^2)^FracPart[p], Int[(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])

Rule 5073

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + (I*n
)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int e^{-\tan ^{-1}(a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx &=\frac{\left (c \sqrt{c+a^2 c x^2}\right ) \int e^{-\tan ^{-1}(a x)} \left (1+a^2 x^2\right )^{3/2} \, dx}{\sqrt{1+a^2 x^2}}\\ &=\frac{\left (c \sqrt{c+a^2 c x^2}\right ) \int (1-i a x)^{\frac{3}{2}-\frac{i}{2}} (1+i a x)^{\frac{3}{2}+\frac{i}{2}} \, dx}{\sqrt{1+a^2 x^2}}\\ &=-\frac{\left (\frac{1}{13}-\frac{5 i}{13}\right ) 2^{\frac{3}{2}+\frac{i}{2}} c (1-i a x)^{\frac{5}{2}-\frac{i}{2}} \sqrt{c+a^2 c x^2} \, _2F_1\left (-\frac{3}{2}-\frac{i}{2},\frac{5}{2}-\frac{i}{2};\frac{7}{2}-\frac{i}{2};\frac{1}{2} (1-i a x)\right )}{a \sqrt{1+a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0222047, size = 98, normalized size = 1. \[ -\frac{\left (\frac{1}{13}-\frac{5 i}{13}\right ) 2^{\frac{3}{2}+\frac{i}{2}} c (1-i a x)^{\frac{5}{2}-\frac{i}{2}} \sqrt{a^2 c x^2+c} \, _2F_1\left (-\frac{3}{2}-\frac{i}{2},\frac{5}{2}-\frac{i}{2};\frac{7}{2}-\frac{i}{2};\frac{1}{2} (1-i a x)\right )}{a \sqrt{a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + a^2*c*x^2)^(3/2)/E^ArcTan[a*x],x]

[Out]

((-1/13 + (5*I)/13)*2^(3/2 + I/2)*c*(1 - I*a*x)^(5/2 - I/2)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[-3/2 - I/2,
5/2 - I/2, 7/2 - I/2, (1 - I*a*x)/2])/(a*Sqrt[1 + a^2*x^2])

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Maple [F]  time = 0.301, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{{\rm e}^{\arctan \left ( ax \right ) }}} \left ({a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(3/2)/exp(arctan(a*x)),x)

[Out]

int((a^2*c*x^2+c)^(3/2)/exp(arctan(a*x)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} e^{\left (-\arctan \left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(3/2)/exp(arctan(a*x)),x, algorithm="maxima")

[Out]

integrate((a^2*c*x^2 + c)^(3/2)*e^(-arctan(a*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} e^{\left (-\arctan \left (a x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(3/2)/exp(arctan(a*x)),x, algorithm="fricas")

[Out]

integral((a^2*c*x^2 + c)^(3/2)*e^(-arctan(a*x)), x)

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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((a**2*c*x**2+c)**(3/2)/exp(atan(a*x)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(3/2)/exp(arctan(a*x)),x, algorithm="giac")

[Out]

Exception raised: TypeError