3.119 \(\int \frac{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=157 \[ -\frac{2 \left (A d^2-B c d+c^2 C\right )}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}} \]

[Out]

-(((I*A + B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f)) - ((B - I*(A - C))*Ar
cTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(c^2*C - B*c*d + A*d^2))/(d*(c^2 + d^2
)*f*Sqrt[c + d*Tan[e + f*x]])

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Rubi [A]  time = 0.293992, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3628, 3539, 3537, 63, 208} \[ -\frac{2 \left (A d^2-B c d+c^2 C\right )}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

-(((I*A + B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f)) - ((B - I*(A - C))*Ar
cTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(c^2*C - B*c*d + A*d^2))/(d*(c^2 + d^2
)*f*Sqrt[c + d*Tan[e + f*x]])

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{A c-c C+B d+(B c-(A-C) d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{c^2+d^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(A-i B-C) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)}+\frac{(A+i B-C) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(i A+B-i C) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d) f}-\frac{(i (A+i B-C)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d) f}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{(A-i B-C) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c-i d) d f}-\frac{(A+i B-C) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c+i d) d f}\\ &=-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{3/2} f}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{3/2} f}-\frac{2 \left (c^2 C-B c d+A d^2\right )}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.92493, size = 218, normalized size = 1.39 \[ \frac{\frac{(d (C-A)+B c) \left ((d-i c) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )+(d+i c) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )\right )}{\left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-i B \left (\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}\right )-\frac{2 C}{\sqrt{c+d \tan (e+f x)}}}{d f} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((-I)*B*(ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]/Sqrt[c - I*d] - ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt
[c + I*d]]/Sqrt[c + I*d]) - (2*C)/Sqrt[c + d*Tan[e + f*x]] + ((B*c + (-A + C)*d)*(((-I)*c + d)*Hypergeometric2
F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] + (I*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*
x])/(c + I*d)]))/((c^2 + d^2)*Sqrt[c + d*Tan[e + f*x]]))/(d*f)

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Maple [B]  time = 0.125, size = 11427, normalized size = 72.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x)

[Out]

result too large to display

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Maxima [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+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral((A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(c + d*tan(e + f*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)/(d*tan(f*x + e) + c)^(3/2), x)