3.1312 \(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=283 \[ \frac{(b c-a d) \sqrt{c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 b f (m+1) (b+i a) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}-\frac{(b c-a d) \sqrt{c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 b f (m+1) (-b+i a) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}} \]

[Out]

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

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Rubi [A]  time = 0.357141, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3575, 912, 137, 136} \[ \frac{(b c-a d) \sqrt{c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 b f (m+1) (b+i a) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}-\frac{(b c-a d) \sqrt{c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 b f (m+1) (-b+i a) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

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

Rubi steps

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

Mathematica [F]  time = 6.81423, size = 0, normalized size = 0. \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.389, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [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((a+b*tan(f*x+e))**m*(c+d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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