3.680 \(\int \frac{1}{\tan ^{\frac{4}{3}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=163 \[ -\frac{3 \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{2 d \sqrt [3]{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{3 \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{2 d \sqrt [3]{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.260991, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3575, 912, 130, 511, 510} \[ -\frac{3 \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{2 d \sqrt [3]{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{3 \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{2 d \sqrt [3]{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(4/3)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

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

Rule 3575

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]

Rule 912

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{1}{\tan ^{\frac{4}{3}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^{4/3} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i}{2 (i-x) x^{4/3} \sqrt{a+b x}}+\frac{i}{2 x^{4/3} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{1}{(i-x) x^{4/3} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{1}{x^{4/3} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (i-x^3\right ) \sqrt{a+b x^3}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (i+x^3\right ) \sqrt{a+b x^3}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d}\\ &=\frac{\left (3 i \sqrt{1+\frac{b \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (i-x^3\right ) \sqrt{1+\frac{b x^3}{a}}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d \sqrt{a+b \tan (c+d x)}}+\frac{\left (3 i \sqrt{1+\frac{b \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (i+x^3\right ) \sqrt{1+\frac{b x^3}{a}}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d \sqrt{a+b \tan (c+d x)}}\\ &=-\frac{3 F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \sqrt{1+\frac{b \tan (c+d x)}{a}}}{2 d \sqrt [3]{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{3 F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \sqrt{1+\frac{b \tan (c+d x)}{a}}}{2 d \sqrt [3]{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 52.0324, size = 23249, normalized size = 142.63 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Tan[c + d*x]^(4/3)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

Result too large to show

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Maple [F]  time = 0.308, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{a+b\tan \left ( dx+c \right ) }}} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(4/3),x)

[Out]

int(1/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^(4/3)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \tan{\left (c + d x \right )}} \tan ^{\frac{4}{3}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))**(1/2)/tan(d*x+c)**(4/3),x)

[Out]

Integral(1/(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**(4/3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^(4/3)), x)