3.2.70 \(\int \frac {(a+b \tan (e+f x))^m (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^2} \, dx\) [170]

Optimal. Leaf size=403 \[ \frac {(A-i B-C) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a+b) (c-i d)^2 f (1+m)}+\frac {(i A-B-i C) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) (c+i d)^2 f (1+m)}-\frac {\left (a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (A d^2 \left (c^2 (2-m)-d^2 m\right )-B c d \left (c^2 (1-m)-d^2 (1+m)\right )-c^2 C \left (c^2 m+d^2 (2+m)\right )\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac {\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.84, antiderivative size = 402, normalized size of antiderivative = 1.00, 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 = {3730, 3734, 3620, 3618, 70, 3715} \begin {gather*} \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+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^4 (-C) m-c^2 C d^2 (m+2)\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 {(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} \end {gather*}

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 70

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

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3715

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 3730

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*Ta
n[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 3734

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]

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 ) \text {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) \text {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)) \text {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*}

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Mathematica [A]
time = 5.70, size = 360, normalized size = 0.89 \begin {gather*} \frac {(a+b \tan (e+f x))^{1+m} \left (-\frac {i \left (\frac {(A-i B-C) (c+i d)^2 (-b c+a d) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac {(A+i B-C) (c-i d)^2 (b c-a d) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right )}{a+i b}\right )}{\left (c^2+d^2\right ) (1+m)}+\frac {2 \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 )}{(b c-a d) \left (c^2+d^2\right ) (1+m)}-\frac {2 \left (c^2 C-B c d+A d^2\right )}{c+d \tan (e+f x)}\right )}{2 (-b c+a d) \left (c^2+d^2\right ) f} \end {gather*}

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

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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m} \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\left (c +d \tan \left (f x +e \right )\right )^{2}}\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right )}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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)

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