3.705 \(\int \frac {\tan ^m(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=179 \[ \frac {\tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {\tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a d (m+1) \sqrt {a+b \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 179, 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 = {3575, 912, 135, 133} \[ \frac {\tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {\tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a d (m+1) \sqrt {a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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[1 + (b*Tan
[c + d*x])/a])/(2*a*d*(1 + m)*Sqrt[a + b*Tan[c + d*x]]) + (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[1 + (b*Tan[c + d*x])/a])/(2*a*d*(1 + m)*Sqrt[a + b*Tan[c + d*x]])

Rule 133

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

Rule 135

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

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Mathematica [F]  time = 19.12, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^m(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx \]

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]

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fricas [F]  time = 2.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m}}{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}, x\right ) \]

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

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

[Out]

Timed out

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maple [F]  time = 0.89, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{m}\left (d x +c \right )}{\left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

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)

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

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

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

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)

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

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

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