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

Optimal. Leaf size=282 \[ \frac{\left (a^2 A+2 a b B-A b^2\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^2}+\frac{b \left (a^2 A b (2-m)+a^3 (-(B-B m))+a b^2 B (m+1)-A b^3 m\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{b \tan (c+d x)}{a}\right )}{a^2 d (m+1) \left (a^2+b^2\right )^2}-\frac{\left (a^2 (-B)+2 a A b+b^2 B\right ) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^2}+\frac{b (A b-a B) \tan ^{m+1}(c+d x)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))} \]

[Out]

((a^2*A - A*b^2 + 2*a*b*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(
(a^2 + b^2)^2*d*(1 + m)) + (b*(a^2*A*b*(2 - m) - A*b^3*m + a*b^2*B*(1 + m) - a^3*(B - B*m))*Hypergeometric2F1[
1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(1 + m))/(a^2*(a^2 + b^2)^2*d*(1 + m)) - ((2*a*A*b - a^2*
B + b^2*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/((a^2 + b^2)^2*d*
(2 + m)) + (b*(A*b - a*B)*Tan[c + d*x]^(1 + m))/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.703814, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3609, 3653, 3538, 3476, 364, 3634, 64} \[ \frac{\left (a^2 A+2 a b B-A b^2\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^2}+\frac{b \left (a^2 A b (2-m)+a^3 (-(B-B m))+a b^2 B (m+1)-A b^3 m\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac{b \tan (c+d x)}{a}\right )}{a^2 d (m+1) \left (a^2+b^2\right )^2}-\frac{\left (a^2 (-B)+2 a A b+b^2 B\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^2}+\frac{b (A b-a B) \tan ^{m+1}(c+d x)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*A - A*b^2 + 2*a*b*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(
(a^2 + b^2)^2*d*(1 + m)) + (b*(a^2*A*b*(2 - m) - A*b^3*m + a*b^2*B*(1 + m) - a^3*(B - B*m))*Hypergeometric2F1[
1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(1 + m))/(a^2*(a^2 + b^2)^2*d*(1 + m)) - ((2*a*A*b - a^2*
B + b^2*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/((a^2 + b^2)^2*d*
(2 + m)) + (b*(A*b - a*B)*Tan[c + d*x]^(1 + m))/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 3609

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*(A*b - a*B)*(a + b*Tan[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[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[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 3538

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

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]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{\tan ^m(c+d x) \left (a^2 A-A b^2 m+a b B (1+m)-a (A b-a B) \tan (c+d x)-b (A b-a B) m \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \tan ^m(c+d x) \left (a \left (a^2 A-A b^2+2 a b B\right )-a \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)\right ) \, dx}{a \left (a^2+b^2\right )^2}+\frac{\left (a^2 b (A b-a B)-a^2 b (A b-a B) m+b^2 \left (a^2 A-A b^2 m+a b B (1+m)\right )\right ) \int \frac{\tan ^m(c+d x) \left (1+\tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )^2}\\ &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (a^2 A-A b^2+2 a b B\right ) \int \tan ^m(c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (2 a A b-a^2 B+b^2 B\right ) \int \tan ^{1+m}(c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (a^2 b (A b-a B)-a^2 b (A b-a B) m+b^2 \left (a^2 A-A b^2 m+a b B (1+m)\right )\right ) \operatorname{Subst}\left (\int \frac{x^m}{a+b x} \, dx,x,\tan (c+d x)\right )}{a \left (a^2+b^2\right )^2 d}\\ &=-\frac{b \left (a^3 B (1-m)-a^2 A b (2-m)+A b^3 m-a b^2 B (1+m)\right ) \, _2F_1\left (1,1+m;2+m;-\frac{b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a^2 \left (a^2+b^2\right )^2 d (1+m)}+\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (a^2 A-A b^2+2 a b B\right ) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right )^2 d}-\frac{\left (2 a A b-a^2 B+b^2 B\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{\left (a^2 A-A b^2+2 a b B\right ) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right )^2 d (1+m)}-\frac{b \left (a^3 B (1-m)-a^2 A b (2-m)+A b^3 m-a b^2 B (1+m)\right ) \, _2F_1\left (1,1+m;2+m;-\frac{b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a^2 \left (a^2+b^2\right )^2 d (1+m)}-\frac{\left (2 a A b-a^2 B+b^2 B\right ) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right )^2 d (2+m)}+\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.75554, size = 239, normalized size = 0.85 \[ \frac{\tan ^{m+1}(c+d x) \left (\frac{a \left (\frac{\left (a^2 A+2 a b B-A b^2\right ) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{m+1}+\frac{\left (a^2 B-2 a A b-b^2 B\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{m+2}\right )}{a^2+b^2}+\frac{b \left (-a^2 A b (m-2)+a^3 B (m-1)+a b^2 B (m+1)-A b^3 m\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{b \tan (c+d x)}{a}\right )}{a (m+1) \left (a^2+b^2\right )}+\frac{b (A b-a B)}{a+b \tan (c+d x)}\right )}{a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[c + d*x]^(1 + m)*((b*(-(a^2*A*b*(-2 + m)) + a^3*B*(-1 + m) - A*b^3*m + a*b^2*B*(1 + m))*Hypergeometric2F1
[1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)])/(a*(a^2 + b^2)*(1 + m)) + (b*(A*b - a*B))/(a + b*Tan[c + d*x]) + (a*
(((a^2*A - A*b^2 + 2*a*b*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2])/(1 + m) + ((-2*a*A*b
+ a^2*B - b^2*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(2 + m)))/(a^2 + b^
2)))/(a*(a^2 + b^2)*d)

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Maple [F]  time = 0.433, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) }{ \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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 )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 ^{m}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**m/(a + b*tan(c + d*x))**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 )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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