3.359 \(\int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=198 \[ -\frac{2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}-\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]

[Out]

((I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(5/2)*d) - ((I*A - B)*ArcTanh[Sqrt[a +
b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(5/2)*d) - (2*a^2*(A*b - a*B))/(3*b^2*(a^2 + b^2)*d*(a + b*Tan[c +
d*x])^(3/2)) + (2*a*(2*A*b^3 - a*(a^2 + 3*b^2)*B))/(b^2*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

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Rubi [A]  time = 0.529017, antiderivative size = 198, 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, 3628, 3539, 3537, 63, 208} \[ -\frac{2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}-\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(5/2)*d) - ((I*A - B)*ArcTanh[Sqrt[a +
b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(5/2)*d) - (2*a^2*(A*b - a*B))/(3*b^2*(a^2 + b^2)*d*(a + b*Tan[c +
d*x])^(3/2)) + (2*a*(2*A*b^3 - a*(a^2 + 3*b^2)*B))/(b^2*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

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 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{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=-\frac{2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\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)}{(a+b \tan (c+d x))^{3/2}} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{-b \left (a^2 A-A b^2+2 a b B\right )+b \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{b \left (a^2+b^2\right )^2}\\ &=-\frac{2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac{(A+i B) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac{2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\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)^2 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)^2 d}\\ &=-\frac{2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\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 )}{b (i a+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 )}{(i a-b)^2 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)^{5/2} d}-\frac{(i A-B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{5/2} d}-\frac{2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.941611, size = 260, normalized size = 1.31 \[ -\frac{b (A b-a B) \left (i (a+i b) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )-(b+i a) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )+3 b B (a+b \tan (c+d x)) \left (i (a+i b) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )-(b+i a) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )+2 (a-i b) (a+i b) (2 a B+A b)+6 b B (a-i b) (a+i b) \tan (c+d x)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

-(2*(a - I*b)*(a + I*b)*(A*b + 2*a*B) + b*(A*b - a*B)*(I*(a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan
[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a + I*b)]) + 6*(a - I
*b)*(a + I*b)*b*B*Tan[c + d*x] + 3*b*B*(I*(a + I*b)*Hypergeometric2F1[-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 + b*Tan[c + d*x]))/(3*b^
2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2))

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Maple [B]  time = 0.102, size = 12849, normalized size = 64.9 \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))^(5/2),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))^(5/2),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))^(5/2),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{5}{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))**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**2/(a + b*tan(c + d*x))**(5/2), 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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^2/(b*tan(d*x + c) + a)^(5/2), x)