3.8.2 \(\int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\) [702]

Optimal. Leaf size=175 \[ \frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]

[Out]

1/2*a*AppellF1(1+m,1,-3/2,2+m,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^(1+m)/d/(1+m)/(
1+b*tan(d*x+c)/a)^(1/2)+1/2*a*AppellF1(1+m,1,-3/2,2+m,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^(1/2)*tan
(d*x+c)^(1+m)/d/(1+m)/(1+b*tan(d*x+c)/a)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3656, 926, 140, 138} \begin {gather*} \frac {a \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} F_1\left (m+1;-\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {a \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} F_1\left (m+1;-\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 138

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

Rule 140

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 926

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 3656

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rubi steps

\begin {align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx &=\frac {\text {Subst}\left (\int \frac {x^m (a+b x)^{3/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {i x^m (a+b x)^{3/2}}{2 (i-x)}+\frac {i x^m (a+b x)^{3/2}}{2 (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i \text {Subst}\left (\int \frac {x^m (a+b x)^{3/2}}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \text {Subst}\left (\int \frac {x^m (a+b x)^{3/2}}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\left (i a \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \left (1+\frac {b x}{a}\right )^{3/2}}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {\left (i a \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \left (1+\frac {b x}{a}\right )^{3/2}}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}}\\ &=\frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 18.93, size = 0, normalized size = 0.00 \begin {gather*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.39, size = 0, normalized size = 0.00 \[\int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(tan(d*x+c)^m*(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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(tan(d*x+c)^m*(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^m, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{m}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(a+b*tan(d*x+c))**(3/2),x)

[Out]

Integral((a + b*tan(c + d*x))**(3/2)*tan(c + d*x)**m, x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (c+d\,x\right )}^m\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________