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

Optimal. Leaf size=196 \[ -\frac{a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{(a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{5/2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d} \]

[Out]

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

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Rubi [A]  time = 0.882202, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3605, 3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{(a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{5/2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3605

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e
+ f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3647

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

Rule 3653

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

Rule 3634

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

Rubi steps

\begin{align*} \int \cot ^2(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}+\int \cot (c+d x) \sqrt{a+b \tan (c+d x)} \left (\frac{1}{2} a (5 A b+2 a B)-\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac{1}{2} b (a A+2 b B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}+2 \int \frac{\cot (c+d x) \left (\frac{1}{4} a^2 (5 A b+2 a B)-\frac{1}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac{1}{4} b \left (a^2 A-2 A b^2-6 a b B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}+2 \int \frac{\frac{1}{2} \left (-a^3 A+3 a A b^2+3 a^2 b B-b^3 B\right )-\frac{1}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} \left (a^2 (5 A b+2 a B)\right ) \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}-\frac{1}{2} \left ((a-i b)^3 (A-i B)\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{1}{2} \left ((a+i b)^3 (A+i B)\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\frac{\left (a^2 (5 A b+2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}+\frac{\left (i (a+i b)^3 (A+i B)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}-\frac{\left ((a-i b)^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac{\left (a^2 (5 A b+2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=-\frac{a^{3/2} (5 A b+2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}+\frac{\left ((a-i b)^3 (A-i B)\right ) \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 d}+\frac{\left ((a+i b)^3 (A+i B)\right ) \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 d}\\ &=-\frac{a^{3/2} (5 A b+2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{(a-i b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{5/2} (i A-B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{b (a A+2 b B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}\\ \end{align*}

Mathematica [B]  time = 1.01344, size = 400, normalized size = 2.04 \[ \frac{2 b B \cot (c+d x) (a+b \tan (c+d x))^{3/2}}{d}+2 \left (-\frac{b (4 a B+A b) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{d}-2 \left (\frac{\left (a^2 A-6 a b B-2 A b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{\frac{i \sqrt{a-i b} \left (-\frac{1}{4} a \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )+\frac{1}{4} i a \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (-a+i b)}-\frac{i \sqrt{a+i b} \left (-\frac{1}{4} a \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )-\frac{1}{4} i a \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (-a-i b)}-\frac{a^{5/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*B*Cot[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/d + 2*(-((b*(A*b + 4*a*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]
])/d) - 2*(-((-(a^(5/2)*(5*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(4*d) + (I*Sqrt[a - I*b]*((
I/4)*a*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B) - (a*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B))/4)*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a - I*b]])/((-a + I*b)*d) - (I*Sqrt[a + I*b]*((-I/4)*a*(3*a^2*A*b - A*b^3 + a^3*B - 3*
a*b^2*B) - (a*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B))/4)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/((-
a - I*b)*d))/a) + ((a^2*A - 2*A*b^2 - 6*a*b*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(4*d)))

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Maple [C]  time = 2.879, size = 88645, normalized size = 452.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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(cot(d*x+c)^2*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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(cot(d*x+c)**2*(a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [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(cot(d*x+c)^2*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out