3.613 \(\int \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=290 \[ \frac{2 \left (15 a^2 A-5 a b B+2 A b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (5 a B+A b) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}+\frac{\sqrt{-b+i a} (A+i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\sqrt{b+i a} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d} \]

[Out]

(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]
]*Sqrt[Tan[c + d*x]])/d - (Sqrt[I*a + b]*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (2*(15*a^2*A + 2*A*b^2 - 5*a*b*B)*Sqrt[Cot[c + d*x]]*Sqrt
[a + b*Tan[c + d*x]])/(15*a^2*d) - (2*(A*b + 5*a*B)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(15*a*d) - (2
*A*Cot[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]])/(5*d)

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Rubi [A]  time = 1.18744, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4241, 3608, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (15 a^2 A-5 a b B+2 A b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (5 a B+A b) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}+\frac{\sqrt{-b+i a} (A+i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\sqrt{b+i a} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]
]*Sqrt[Tan[c + d*x]])/d - (Sqrt[I*a + b]*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (2*(15*a^2*A + 2*A*b^2 - 5*a*b*B)*Sqrt[Cot[c + d*x]]*Sqrt
[a + b*Tan[c + d*x]])/(15*a^2*d) - (2*(A*b + 5*a*B)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(15*a*d) - (2
*A*Cot[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]])/(5*d)

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3608

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

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3616

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

Rule 3615

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}-\frac{1}{5} \left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{1}{2} (-A b-5 a B)+\frac{5}{2} (a A-b B) \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 (A b+5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}+\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{1}{4} \left (-15 a^2 A-2 A b^2+5 a b B\right )-\frac{15}{4} a (A b+a B) \tan (c+d x)-\frac{1}{2} b (A b+5 a B) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{15 a}\\ &=\frac{2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (A b+5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}-\frac{\left (8 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{15}{8} a^2 (A b+a B)-\frac{15}{8} a^2 (a A-b B) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (A b+5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}-\frac{\left (4 \left (\frac{15}{8} a^2 (A b+a B)-\frac{15}{8} i a^2 (a A-b B)\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}-\frac{\left (4 \left (\frac{15}{8} a^2 (A b+a B)+\frac{15}{8} i a^2 (a A-b B)\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (A b+5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}-\frac{\left (4 \left (\frac{15}{8} a^2 (A b+a B)-\frac{15}{8} i a^2 (a A-b B)\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d}-\frac{\left (4 \left (\frac{15}{8} a^2 (A b+a B)+\frac{15}{8} i a^2 (a A-b B)\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d}\\ &=\frac{2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (A b+5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}-\frac{\left (8 \left (\frac{15}{8} a^2 (A b+a B)-\frac{15}{8} i a^2 (a A-b B)\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{15 a^2 d}-\frac{\left (8 \left (\frac{15}{8} a^2 (A b+a B)+\frac{15}{8} i a^2 (a A-b B)\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{15 a^2 d}\\ &=\frac{\sqrt{i a-b} (A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}-\frac{\sqrt{i a+b} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 (A b+5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 d}\\ \end{align*}

Mathematica [A]  time = 3.98638, size = 251, normalized size = 0.87 \[ -\frac{\cot ^{\frac{5}{2}}(c+d x) \left (2 \sqrt{a+b \tan (c+d x)} \left (\left (-15 a^2 A+5 a b B-2 A b^2\right ) \tan ^2(c+d x)+3 a^2 A+a (5 a B+A b) \tan (c+d x)\right )+15 \sqrt [4]{-1} a^2 \sqrt{-a-i b} (A+i B) \tan ^{\frac{5}{2}}(c+d x) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+15 \sqrt [4]{-1} a^2 \sqrt{a-i b} (A-i B) \tan ^{\frac{5}{2}}(c+d x) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )}{15 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

-(Cot[c + d*x]^(5/2)*(15*(-1)^(1/4)*a^2*Sqrt[-a - I*b]*(A + I*B)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c
+ d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(5/2) + 15*(-1)^(1/4)*a^2*Sqrt[a - I*b]*(A - I*B)*ArcTanh[((-
1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(5/2) + 2*Sqrt[a + b*Tan[c +
d*x]]*(3*a^2*A + a*(A*b + 5*a*B)*Tan[c + d*x] + (-15*a^2*A - 2*A*b^2 + 5*a*b*B)*Tan[c + d*x]^2)))/(15*a^2*d)

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Maple [C]  time = 1.63, size = 42569, normalized size = 146.8 \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)^(7/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*cot(d*x + c)^(7/2), x)

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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)^(7/2)*(a+b*tan(d*x+c))^(1/2)*(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)**(7/2)*(a+b*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*cot(d*x + c)^(7/2), x)