3.661 \(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=173 \[ \frac {(A+i B) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d}+\frac {(A-i B) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d} \]

[Out]

1/5*(A+I*B)*AppellF1(5/2,1,-n,7/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^n/d/((1+b*t
an(d*x+c)/a)^n)+1/5*(A-I*B)*AppellF1(5/2,1,-n,7/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(5/2)*(a+b*tan(d*x+
c))^n/d/((1+b*tan(d*x+c)/a)^n)

________________________________________________________________________________________

Rubi [A]  time = 0.37, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3603, 3602, 130, 511, 510} \[ \frac {(A+i B) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d}+\frac {(A-i B) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

((A + I*B)*AppellF1[5/2, 1, -n, 7/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(5/2)*(a + b*Tan[c
+ d*x])^n)/(5*d*(1 + (b*Tan[c + d*x])/a)^n) + ((A - I*B)*AppellF1[5/2, 1, -n, 7/2, I*Tan[c + d*x], -((b*Tan[c
+ d*x])/a)]*Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^n)/(5*d*(1 + (b*Tan[c + d*x])/a)^n)

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 510

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

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 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 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac {1}{2} (A-i B) \int (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \, dx+\frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \, dx\\ &=\frac {(A-i B) \operatorname {Subst}\left (\int \frac {x^{3/2} (a+b x)^n}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(A+i B) \operatorname {Subst}\left (\int \frac {x^{3/2} (a+b x)^n}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {(A-i B) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^n}{1-i x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {(A+i B) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^n}{1+i x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {\left ((A-i B) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1+\frac {b x^2}{a}\right )^n}{1-i x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left ((A+i B) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1+\frac {b x^2}{a}\right )^n}{1+i x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(A+i B) F_1\left (\frac {5}{2};1,-n;\frac {7}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{5 d}+\frac {(A-i B) F_1\left (\frac {5}{2};1,-n;\frac {7}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 2.39, size = 0, normalized size = 0.00 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

Integrate[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]), x]

________________________________________________________________________________________

fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \tan \left (d x + c\right )^{2} + A \tan \left (d x + c\right )\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [F]  time = 1.34, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*tan(d*x + c)^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n, x)

________________________________________________________________________________________

sympy [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)**(3/2)*(a+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________