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

Optimal. Leaf size=328 \[ -\frac {\left (\sqrt {-b^2} (A-C)+b B\right ) \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac {1}{2};1,-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {b \tan (c+d x)}{a}+1\right )}{b d \left (a-\sqrt {-b^2}\right )}-\frac {\left (b B-\sqrt {-b^2} (A-C)\right ) \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac {1}{2};1,-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}},\frac {b \tan (c+d x)}{a}+1\right )}{b d \left (a+\sqrt {-b^2}\right )}+\frac {2 C \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {b \tan (c+d x)}{a}+1\right )}{b d} \]

[Out]

2*C*hypergeom([1/2, -m],[3/2],1+b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^m/b/d/((-b*tan(d*x+c)/a)^m)-
AppellF1(1/2,1,-m,3/2,(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)),1+b*tan(d*x+c)/a)*(b*B-(A-C)*(-b^2)^(1/2))*(a+b*tan(d*
x+c))^(1/2)*tan(d*x+c)^m/b/d/(a+(-b^2)^(1/2))/((-b*tan(d*x+c)/a)^m)-AppellF1(1/2,1,-m,3/2,(a+b*tan(d*x+c))/(a-
(-b^2)^(1/2)),1+b*tan(d*x+c)/a)*(b*B+(A-C)*(-b^2)^(1/2))*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^m/b/d/(a-(-b^2)^(1/
2))/((-b*tan(d*x+c)/a)^m)

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Rubi [A]  time = 1.56, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3655, 6720, 1692, 246, 245, 430, 429} \[ -\frac {\left (\sqrt {-b^2} (A-C)+b B\right ) \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac {1}{2};1,-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {b \tan (c+d x)}{a}+1\right )}{b d \left (a-\sqrt {-b^2}\right )}-\frac {\left (b B-\sqrt {-b^2} (A-C)\right ) \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac {1}{2};1,-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}},\frac {b \tan (c+d x)}{a}+1\right )}{b d \left (a+\sqrt {-b^2}\right )}+\frac {2 C \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {b \tan (c+d x)}{a}+1\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*B + Sqrt[-b^2]*(A - C))*AppellF1[1/2, 1, -m, 3/2, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]), 1 + (b*Tan[c +
d*x])/a]*Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*(a - Sqrt[-b^2])*d*(-((b*Tan[c + d*x])/a))^m)) - ((b*B -
Sqrt[-b^2]*(A - C))*AppellF1[1/2, 1, -m, 3/2, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2]), 1 + (b*Tan[c + d*x])/a]*T
an[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*(a + Sqrt[-b^2])*d*(-((b*Tan[c + d*x])/a))^m) + (2*C*Hypergeometric
2F1[1/2, -m, 3/2, 1 + (b*Tan[c + d*x])/a]*Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*d*(-((b*Tan[c + d*x])/a)
)^m)

Rule 245

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 3655

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] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^m \left (A+B x+C x^2\right )}{\sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (\frac {-a+x^2}{b}\right )^m \left (A b^2+\left (a-x^2\right ) \left (-b B+C \left (a-x^2\right )\right )\right )}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (-a+x^2\right )^m \left (A b^2+\left (a-x^2\right ) \left (-b B+C \left (a-x^2\right )\right )\right )}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \left (C \left (-a+x^2\right )^m+\frac {\left (-a+x^2\right )^m \left (b (A b-a B-b C)+b B x^2\right )}{a^2+b^2-2 a x^2+x^4}\right ) \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (-a+x^2\right )^m \left (b (A b-a B-b C)+b B x^2\right )}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {\left (2 C \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \left (-a+x^2\right )^m \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \left (\frac {\left (b B-\sqrt {-b^2} (A-C)\right ) \left (-a+x^2\right )^m}{-2 a-2 \sqrt {-b^2}+2 x^2}+\frac {\left (b B+\sqrt {-b^2} (A-C)\right ) \left (-a+x^2\right )^m}{-2 a+2 \sqrt {-b^2}+2 x^2}\right ) \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {\left (2 C \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m}\right ) \operatorname {Subst}\left (\int \left (1-\frac {x^2}{a}\right )^m \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {2 C \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{b d}+\frac {\left (2 \left (b B-\sqrt {-b^2} (A-C)\right ) \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (-a+x^2\right )^m}{-2 a-2 \sqrt {-b^2}+2 x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {\left (2 \left (b B+\sqrt {-b^2} (A-C)\right ) \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (-a+x^2\right )^m}{-2 a+2 \sqrt {-b^2}+2 x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {2 C \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{b d}+\frac {\left (2 \left (b B-\sqrt {-b^2} (A-C)\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x^2}{a}\right )^m}{-2 a-2 \sqrt {-b^2}+2 x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {\left (2 \left (b B+\sqrt {-b^2} (A-C)\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x^2}{a}\right )^m}{-2 a+2 \sqrt {-b^2}+2 x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=-\frac {\left (b B+\sqrt {-b^2} (A-C)\right ) F_1\left (\frac {1}{2};1,-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{b \left (a-\sqrt {-b^2}\right ) d}-\frac {\left (b B-\sqrt {-b^2} (A-C)\right ) F_1\left (\frac {1}{2};1,-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{b \left (a+\sqrt {-b^2}\right ) d}+\frac {2 C \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{b d}\\ \end {align*}

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Mathematica [F]  time = 28.56, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{\sqrt {b \tan \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 2.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )+C \left (\tan ^{2}\left (d x +c \right )\right )\right )}{\sqrt {a +b \tan \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (C\,{\mathrm {tan}\left (c+d\,x\right )}^2+B\,\mathrm {tan}\left (c+d\,x\right )+A\right )}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(A+B*tan(d*x+c)+C*tan(d*x+c)**2)/(a+b*tan(d*x+c))**(1/2),x)

[Out]

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

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