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

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

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Rubi [A]  time = 0.07, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {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[E^ArcTan[a*x]*(c + a^2*c*x^2)^(3/2),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 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]))

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 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])

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*}

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Mathematica [A]  time = 0.02, size = 98, normalized size = 1.00 \[ \frac {\left (\frac {1}{13}+\frac {5 i}{13}\right ) 2^{\frac {3}{2}-\frac {i}{2}} c (1-i a x)^{\frac {5}{2}+\frac {i}{2}} \sqrt {a^2 c x^2+c} \, _2F_1\left (-\frac {3}{2}+\frac {i}{2},\frac {5}{2}+\frac {i}{2};\frac {7}{2}+\frac {i}{2};\frac {1}{2} (1-i a x)\right )}{a \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTan[a*x]*(c + a^2*c*x^2)^(3/2),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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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} e^{\left (\arctan \left (a x\right )\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arctan \left (a x \right )} \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} e^{\left (\arctan \left (a x\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} e^{\operatorname {atan}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(a**2*x**2 + 1))**(3/2)*exp(atan(a*x)), x)

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