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

Optimal. Leaf size=163 \[ \frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (\frac {7}{3};1,\frac {1}{2};\frac {10}{3};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (\frac {7}{3};1,\frac {1}{2};\frac {10}{3};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 d \sqrt {a+b \tan (c+d x)}} \]

[Out]

3/14*AppellF1(7/3,1,1/2,10/3,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x+c)/a)^(1/2)*tan(d*x+c)^(7/3)/d/(a+b*t
an(d*x+c))^(1/2)+3/14*AppellF1(7/3,1,1/2,10/3,I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x+c)/a)^(1/2)*tan(d*x+c
)^(7/3)/d/(a+b*tan(d*x+c))^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3575, 912, 130, 511, 510} \[ \frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (\frac {7}{3};1,\frac {1}{2};\frac {10}{3};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (\frac {7}{3};1,\frac {1}{2};\frac {10}{3};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 d \sqrt {a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[7/3, 1, 1/2, 10/3, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(7/3)*Sqrt[1 + (b*Tan[c
+ d*x])/a])/(14*d*Sqrt[a + b*Tan[c + d*x]]) + (3*AppellF1[7/3, 1, 1/2, 10/3, I*Tan[c + d*x], -((b*Tan[c + d*x]
)/a)]*Tan[c + d*x]^(7/3)*Sqrt[1 + (b*Tan[c + d*x])/a])/(14*d*Sqrt[a + b*Tan[c + d*x]])

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 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 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]

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {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 x^{4/3}}{2 (i-x) \sqrt {a+b x}}+\frac {i x^{4/3}}{2 (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i \operatorname {Subst}\left (\int \frac {x^{4/3}}{(i-x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \operatorname {Subst}\left (\int \frac {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 {x^6}{\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 {x^6}{\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 {x^6}{\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 {x^6}{\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 {7}{3};1,\frac {1}{2};\frac {10}{3};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 F_1\left (\frac {7}{3};1,\frac {1}{2};\frac {10}{3};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}

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Mathematica [B]  time = 26.82, size = 21046, normalized size = 129.12 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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

[Out]

Timed out

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

[Out]

Timed out

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maple [F]  time = 2.01, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{\frac {4}{3}}\left (d x +c \right )}{\sqrt {a +b \tan \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{\frac {4}{3}}}{\sqrt {b \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{4/3}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{\frac {4}{3}}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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