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

Optimal. Leaf size=167 \[ -\frac{2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{3/2}}-\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{3/2}}+\frac{2 B \sqrt{a+b \tan (c+d x)}}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.44311, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3604, 3630, 3539, 3537, 63, 208} \[ -\frac{2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{3/2}}-\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{3/2}}+\frac{2 B \sqrt{a+b \tan (c+d x)}}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3604

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((B*c - A*d)*(b*c - a*d)^2*(c + d*Tan[e + f*x])^(n + 1))/(f*d^2*(n +
1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a^2
*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^
2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^
2, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rule 3630

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

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

Mathematica [C]  time = 1.27118, size = 248, normalized size = 1.49 \[ \frac{\frac{(A b-a B) \left ((b-i a) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )+(b+i a) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )}{\left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{4 a B-2 A b}{b \sqrt{a+b \tan (c+d x)}}+\frac{2 B \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}}+i B \left (\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{\sqrt{a-i b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{\sqrt{a+i b}}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*B*(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/Sqrt[a - I*b] - ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a
+ I*b]]/Sqrt[a + I*b]) + (-2*A*b + 4*a*B)/(b*Sqrt[a + b*Tan[c + d*x]]) + ((A*b - a*B)*(((-I)*a + b)*Hypergeome
tric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] + (I*a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c
+ d*x])/(a + I*b)]))/((a^2 + b^2)*Sqrt[a + b*Tan[c + d*x]]) + (2*B*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]])/(b
*d)

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Maple [B]  time = 0.115, size = 7982, normalized size = 47.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(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(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(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(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(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{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^2/(b*tan(d*x + c) + a)^(3/2), x)