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

Optimal. Leaf size=403 \[ -\frac{(a+b \tan (e+f x))^{m+1} \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )-b \left (A d^2 \left (c^2 (2-m)-d^2 m\right )-B c d \left (c^2 (1-m)-d^2 (m+1)\right )+c^2 (-C) \left (c^2 m+d^2 (m+2)\right )\right )\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac{(A-i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^2}+\frac{(i A-B-i C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (c+i d)^2}+\frac{\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))} \]

[Out]

((A - I*B - C)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m)
)/(2*(I*a + b)*(c - I*d)^2*f*(1 + m)) + ((I*A - B - I*C)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x
])/(a + I*b)]*(a + b*Tan[e + f*x])^(1 + m))/(2*(a + I*b)*(c + I*d)^2*f*(1 + m)) - ((a*d^2*(2*c*(A - C)*d - B*(
c^2 - d^2)) - b*(A*d^2*(c^2*(2 - m) - d^2*m) - B*c*d*(c^2*(1 - m) - d^2*(1 + m)) - c^2*C*(c^2*m + d^2*(2 + m))
))*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d))]*(a + b*Tan[e + f*x])^(1 + m))/(
(b*c - a*d)^2*(c^2 + d^2)^2*f*(1 + m)) + ((c^2*C - B*c*d + A*d^2)*(a + b*Tan[e + f*x])^(1 + m))/((b*c - a*d)*(
c^2 + d^2)*f*(c + d*Tan[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.21502, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3649, 3653, 3539, 3537, 68, 3634} \[ -\frac{(a+b \tan (e+f x))^{m+1} \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )-b \left (A c^2 d^2 (2-m)-A d^4 m-B \left (c^3 d (1-m)-c d^3 (m+1)\right )-c^2 C d^2 (m+2)+c^4 (-C) m\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac{\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))}+\frac{(A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^2}+\frac{(i A-B-i C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (c+i d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

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

Mathematica [A]  time = 6.18405, size = 563, normalized size = 1.4 \[ -\frac{\left (A d^2-c (B d-c C)\right ) (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (a d-b c) (c+d \tan (e+f x))}-\frac{-\frac{(a+b \tan (e+f x))^{m+1} \left (d^2 \left ((c C-B d) (a d-b c (m+1))-A \left (a c d-b \left (c^2-d^2 m\right )\right )\right )-c d (b c-a d) (B c-d (A-C))-b c^2 m \left (A d^2-B c d+c^2 C\right )\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{d (a+b \tan (e+f x))}{a d-b c}\right )}{f (m+1) \left (c^2+d^2\right ) (a d-b c)}+\frac{\frac{i \left (-(b c-a d) \left (-A \left (c^2-d^2\right )-2 B c d+c^2 C-C d^2\right )-i (b c-a d) \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{-i a-i b \tan (e+f x)}{b-i a}\right )}{2 f (m+1) (a+i b)}-\frac{i \left (-(b c-a d) \left (-A \left (c^2-d^2\right )-2 B c d+c^2 C-C d^2\right )+i (b c-a d) \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{i a+i b \tan (e+f x)}{-b-i a}\right )}{2 f (m+1) (a-i b)}}{c^2+d^2}}{\left (c^2+d^2\right ) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A*d^2 - c*(-(c*C) + B*d))*(a + b*Tan[e + f*x])^(1 + m))/((-(b*c) + a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])
)) - (-(((-(c*d*(b*c - a*d)*(B*c - (A - C)*d)) - b*c^2*(c^2*C - B*c*d + A*d^2)*m + d^2*((c*C - B*d)*(a*d - b*c
*(1 + m)) - A*(a*c*d - b*(c^2 - d^2*m))))*Hypergeometric2F1[1, 1 + m, 2 + m, (d*(a + b*Tan[e + f*x]))/(-(b*c)
+ a*d)]*(a + b*Tan[e + f*x])^(1 + m))/((-(b*c) + a*d)*(c^2 + d^2)*f*(1 + m))) + (((I/2)*(-((b*c - a*d)*(c^2*C
- 2*B*c*d - C*d^2 - A*(c^2 - d^2))) - I*(b*c - a*d)*(2*c*(A - C)*d - B*(c^2 - d^2)))*Hypergeometric2F1[1, 1 +
m, 2 + m, ((-I)*a - I*b*Tan[e + f*x])/((-I)*a + b)]*(a + b*Tan[e + f*x])^(1 + m))/((a + I*b)*f*(1 + m)) - ((I/
2)*(-((b*c - a*d)*(c^2*C - 2*B*c*d - C*d^2 - A*(c^2 - d^2))) + I*(b*c - a*d)*(2*c*(A - C)*d - B*(c^2 - d^2)))*
Hypergeometric2F1[1, 1 + m, 2 + m, -((I*a + I*b*Tan[e + f*x])/((-I)*a - b))]*(a + b*Tan[e + f*x])^(1 + m))/((a
- I*b)*f*(1 + m)))/(c^2 + d^2))/((-(b*c) + a*d)*(c^2 + d^2))

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Maple [F]  time = 0.641, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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