3.12.86 \(\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx\) [1186]

Optimal. Leaf size=123 \[ -\frac {(i c-d) F_1\left (m;-\frac {3}{2},1;1+m;-\frac {d (1+i \tan (e+f x))}{i c-d},\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m \sqrt {c+d \tan (e+f x)}}{2 f m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3645, 142, 141} \begin {gather*} -\frac {(-d+i c) (a+i a \tan (e+f x))^m \sqrt {c+d \tan (e+f x)} F_1\left (m;-\frac {3}{2},1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 142

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 3645

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dis
t[a*(b/f), Subst[Int[(a + x)^(m - 1)*((c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b,
c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rubi steps

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

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Mathematica [F]
time = 27.04, size = 0, normalized size = 0.00 \begin {gather*} \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 3.00, size = 0, normalized size = 0.00 \[\int \left (a +i a \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integral(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*sqrt(
((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))/(e^(2*I*f*x + 2*I*e) + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*(tan(e + f*x) - I))**m*(c + d*tan(e + f*x))**(3/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^m*(c + d*tan(e + f*x))^(3/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^m*(c + d*tan(e + f*x))^(3/2), x)

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