3.23 \(\int \frac {\tan (d+e x)}{(a+b \tan (d+e x)+c \tan ^2(d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=635 \[ \frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \tanh ^{-1}\left (\frac {-b \left (-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \left (a^2-2 a c+b^2+c^2\right )^{3/2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \tanh ^{-1}\left (\frac {-b \left (\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (-\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \left (a^2-2 a c+b^2+c^2\right )^{3/2}}+\frac {2 \left (a \left (b^2-2 c (a-c)\right )+b c (a+c) \tan (d+e x)\right )}{e \left ((a-c)^2+b^2\right ) \left (b^2-4 a c\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \]

[Out]

-1/2*arctanh(1/2*(b^2-(a-c)*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))-b*(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*
2^(1/2)/(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-b^2-2*a*c+c^2-(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a
+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))*(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)*(a^2-b^2-2*a*c+c^2-(a-c)*(a^2-2
*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-2*a*c+b^2+c^2)^(3/2)/e*2^(1/2)+1/2*arctanh(1/2*(b^2-(a-c)*(a-c+(a^2-2*a*c+b^2+
c^2)^(1/2))-b*(2*a-2*c-(a^2-2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*2^(1/2)/(2*a-2*c-(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2
)/(a^2-b^2-2*a*c+c^2+(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))*(2*a-2*c-(a
^2-2*a*c+b^2+c^2)^(1/2))^(1/2)*(a^2-b^2-2*a*c+c^2+(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-2*a*c+b^2+c^2)^(
3/2)/e*2^(1/2)+2*(a*(b^2-2*(a-c)*c)+b*c*(a+c)*tan(e*x+d))/(b^2+(a-c)^2)/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)+c*tan(e
*x+d)^2)^(1/2)

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Rubi [A]  time = 3.79, antiderivative size = 635, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3700, 1018, 1036, 1030, 208} \[ \frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \tanh ^{-1}\left (\frac {-b \left (-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \left (a^2-2 a c+b^2+c^2\right )^{3/2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \tanh ^{-1}\left (\frac {-b \left (\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (-\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \left (a^2-2 a c+b^2+c^2\right )^{3/2}}+\frac {2 \left (a \left (b^2-2 c (a-c)\right )+b c (a+c) \tan (d+e x)\right )}{e \left ((a-c)^2+b^2\right ) \left (b^2-4 a c\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2*a - 2*c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 + (a - c)*Sqrt[a^2 + b^2 - 2*a*c
+ c^2]]*ArcTanh[(b^2 - (a - c)*(a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]) - b*(2*a - 2*c - Sqrt[a^2 + b^2 - 2*a*
c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[2*a - 2*c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2
+ (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*(a^2 + b^2 -
2*a*c + c^2)^(3/2)*e) - (Sqrt[2*a - 2*c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 - (a - c
)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh[(b^2 - (a - c)*(a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]) - b*(2*a - 2*c
+ Sqrt[a^2 + b^2 - 2*a*c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[2*a - 2*c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[
a^2 - b^2 - 2*a*c + c^2 - (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])]
)/(Sqrt[2]*(a^2 + b^2 - 2*a*c + c^2)^(3/2)*e) + (2*(a*(b^2 - 2*(a - c)*c) + b*c*(a + c)*Tan[d + e*x]))/((b^2 +
(a - c)^2)*(b^2 - 4*a*c)*e*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])

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]

Rule 1018

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^(q + 1)*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*(
2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(2*a*f)) - h*(b*c*d + a*b*f))*x))/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)
*(p + 1)), x] + Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + f
*x^2)^q*Simp[(b*h - 2*g*c)*((c*d - a*f)^2 - (b*d)*(-(b*f)))*(p + 1) + (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*
(c*d - a*f)))*(a*f*(p + 1) - c*d*(p + 2)) - (2*f*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*
(2*a*f)))*(p + q + 2) - (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(b*f*(p + 1)))*x - c*f*(b^2*(g*f
) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}
, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1
])

Rule 1030

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
a*g*h, Subst[Int[1/Simp[2*a^2*g*h*c + a*e*x^2, x], x], x, Simp[a*h - g*c*x, x]/Sqrt[d + e*x + f*x^2]], x] /; F
reeQ[{a, c, d, e, f, g, h}, x] && EqQ[a*h^2*e + 2*g*h*(c*d - a*f) - g^2*c*e, 0]

Rule 1036

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(c*d - a*f)^2 + a*c*e^2, 2]}, Dist[1/(2*q), Int[Simp[-(a*h*e) - g*(c*d - a*f - q) + (h*(c*d - a*f + q) -
g*c*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[Simp[-(a*h*e) - g*(c*d - a*f + q
) + (h*(c*d - a*f - q) - g*c*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g,
h}, x] && NeQ[e^2 - 4*d*f, 0] && NegQ[-(a*c)]

Rule 3700

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[((x/f)^m*(a + b*x^n + c*x^(2*n))^p)/(f^2 + x^2
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [C]  time = 4.75, size = 318, normalized size = 0.50 \[ \frac {\frac {\left (4 a^2 c-a \left (b^2-4 i b c+4 c^2\right )+b^2 (c-i b)\right ) \tanh ^{-1}\left (\frac {2 a+(b-2 i c) \tan (d+e x)-i b}{2 \sqrt {a-i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {a-i b-c}}+\frac {\left (4 a^2 c-a \left (b^2+4 i b c+4 c^2\right )+b^2 (c+i b)\right ) \tanh ^{-1}\left (\frac {2 a+(b+2 i c) \tan (d+e x)+i b}{2 \sqrt {a+i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {a+i b-c}}+\frac {4 \left (a \left (2 c (c-a)+b^2\right )+b c (a+c) \tan (d+e x)\right )}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}}{2 e \left ((a-c)^2+b^2\right ) \left (b^2-4 a c\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((4*a^2*c + b^2*((-I)*b + c) - a*(b^2 - (4*I)*b*c + 4*c^2))*ArcTanh[(2*a - I*b + (b - (2*I)*c)*Tan[d + e*x])/
(2*Sqrt[a - I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/Sqrt[a - I*b - c] + ((4*a^2*c + b^2*(I*b +
c) - a*(b^2 + (4*I)*b*c + 4*c^2))*ArcTanh[(2*a + I*b + (b + (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a + I*b - c]*Sqrt[
a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/Sqrt[a + I*b - c] + (4*(a*(b^2 + 2*c*(-a + c)) + b*c*(a + c)*Tan[d +
e*x]))/Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(2*(b^2 + (a - c)^2)*(b^2 - 4*a*c)*e)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^2-1)]Discontinuities at zeroes of t_nostep^2-1 were not checkedWarning, integration of abs or sign
assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time:
8.39Not invertible Error: Bad Argument Value

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maple [B]  time = 0.80, size = 13066297, normalized size = 20576.85 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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