3.8.16 \(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx\) [716]

Optimal. Leaf size=159 \[ \frac {F_1\left (\frac {3}{2};1,-n;\frac {5}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{3 d}+\frac {F_1\left (\frac {3}{2};1,-n;\frac {5}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{3 d} \]

[Out]

1/3*AppellF1(3/2,1,-n,5/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n/d/((1+b*tan(d*x+c
)/a)^n)+1/3*AppellF1(3/2,1,-n,5/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n/d/((1+b*ta
n(d*x+c)/a)^n)

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Rubi [A]
time = 0.16, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3656, 926, 129, 525, 524} \begin {gather*} \frac {\tan ^{\frac {3}{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 {3}{2};1,-n;\frac {5}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d}+\frac {\tan ^{\frac {3}{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 {3}{2};1,-n;\frac {5}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n,x]

[Out]

(AppellF1[3/2, 1, -n, 5/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^n
)/(3*d*(1 + (b*Tan[c + d*x])/a)^n) + (AppellF1[3/2, 1, -n, 5/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c +
d*x]^(3/2)*(a + b*Tan[c + d*x])^n)/(3*d*(1 + (b*Tan[c + d*x])/a)^n)

Rule 129

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 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 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx &=\frac {\text {Subst}\left (\int \frac {\sqrt {x} (a+b x)^n}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {i \sqrt {x} (a+b x)^n}{2 (i-x)}+\frac {i \sqrt {x} (a+b x)^n}{2 (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i \text {Subst}\left (\int \frac {\sqrt {x} (a+b x)^n}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \text {Subst}\left (\int \frac {\sqrt {x} (a+b x)^n}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {i \text {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^n}{i-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {i \text {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^n}{i+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {\left (i (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}\right ) \text {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^n}{i-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left (i (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}\right ) \text {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^n}{i+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {F_1\left (\frac {3}{2};1,-n;\frac {5}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{3 d}+\frac {F_1\left (\frac {3}{2};1,-n;\frac {5}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{3 d}\\ \end {align*}

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Mathematica [F]
time = 1.63, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n,x]

[Out]

Integrate[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n, x]

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Maple [F]
time = 0.43, size = 0, normalized size = 0.00 \[\int \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n,x)

[Out]

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n,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(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*sqrt(tan(d*x + c)), 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(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*sqrt(tan(c + d*x)), x)

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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(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*sqrt(tan(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^n, x)

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