3.492 \(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=193 \[ \frac {(A+i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {5}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a^2 d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {5}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a^2 d (m+1) \sqrt {a+b \tan (c+d x)}} \]

[Out]

1/2*(A+I*B)*AppellF1(1+m,1,5/2,2+m,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x+c)/a)^(1/2)*tan(d*x+c)^(1+m)/a^
2/d/(1+m)/(a+b*tan(d*x+c))^(1/2)+1/2*(A-I*B)*AppellF1(1+m,1,5/2,2+m,I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x
+c)/a)^(1/2)*tan(d*x+c)^(1+m)/a^2/d/(1+m)/(a+b*tan(d*x+c))^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3603, 3602, 135, 133} \[ \frac {(A+i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {5}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a^2 d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {5}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a^2 d (m+1) \sqrt {a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((A + I*B)*AppellF1[1 + m, 5/2, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[
1 + (b*Tan[c + d*x])/a])/(2*a^2*d*(1 + m)*Sqrt[a + b*Tan[c + d*x]]) + ((A - I*B)*AppellF1[1 + m, 5/2, 1, 2 + m
, -((b*Tan[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[1 + (b*Tan[c + d*x])/a])/(2*a^2*d*(1 + m)*S
qrt[a + b*Tan[c + d*x]])

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 3602

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] &&  !IntegerQ
[m] &&  !IntegerQ[n] &&  !IntegersQ[2*m, 2*n] && EqQ[A^2 + B^2, 0]

Rule 3603

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] &&  !Integ
erQ[m] &&  !IntegerQ[n] &&  !IntegersQ[2*m, 2*n] && NeQ[A^2 + B^2, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=\frac {1}{2} (A-i B) \int \frac {(1+i \tan (c+d x)) \tan ^m(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx+\frac {1}{2} (A+i B) \int \frac {(1-i \tan (c+d x)) \tan ^m(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\\ &=\frac {(A-i B) \operatorname {Subst}\left (\int \frac {x^m}{(1-i x) (a+b x)^{5/2}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(A+i B) \operatorname {Subst}\left (\int \frac {x^m}{(1+i x) (a+b x)^{5/2}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\left ((A-i B) \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {x^m}{(1-i x) \left (1+\frac {b x}{a}\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{2 a^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left ((A+i B) \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {x^m}{(1+i x) \left (1+\frac {b x}{a}\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{2 a^2 d \sqrt {a+b \tan (c+d x)}}\\ &=\frac {(A+i B) F_1\left (1+m;\frac {5}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 a^2 d (1+m) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) F_1\left (1+m;\frac {5}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 a^2 d (1+m) \sqrt {a+b \tan (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 76.75, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

Integrate[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2), x]

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m}}{b^{3} \tan \left (d x + c\right )^{3} + 3 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^m/(b^3*tan(d*x + c)^3 + 3*a*b^2*tan(d*x +
c)^2 + 3*a^2*b*tan(d*x + c) + a^3), 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))/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 1.45, size = 0, normalized size = 0.00 \[ \int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx \]

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))^(5/2),x)

[Out]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^m/(b*tan(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(5/2),x)

[Out]

int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

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))**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**m/(a + b*tan(c + d*x))**(5/2), x)

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