3.1008 \(\int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=159 \[ -\frac{i a^{3/2} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{a c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \]

[Out]

((-I)*a^(3/2)*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f + (
a*c^2*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*f) - ((I/3)*c*(a + I*a*Tan[e + f*
x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/f

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Rubi [A]  time = 0.161152, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3523, 49, 38, 63, 217, 203} \[ -\frac{i a^{3/2} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{a c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-I)*a^(3/2)*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f + (
a*c^2*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*f) - ((I/3)*c*(a + I*a*Tan[e + f*
x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/f

Rule 3523

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

Rule 49

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*(m
+ n + 1)), x] + Dist[(2*c*n)/(m + n + 1), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \sqrt{a+i a x} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac{\left (a c^2\right ) \operatorname{Subst}\left (\int \sqrt{a+i a x} \sqrt{c-i c x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac{\left (a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{a c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac{\left (i a c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{f}\\ &=\frac{a c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac{\left (i a c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c-i c \tan (e+f x)}}\right )}{f}\\ &=-\frac{i a^{3/2} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{a c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 4.09308, size = 169, normalized size = 1.06 \[ -\frac{i e^{-2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^{5/2} \left (e^{i (e+f x)} \left (8 e^{2 i (e+f x)}+3 e^{4 i (e+f x)}-3\right )+3 \left (1+e^{2 i (e+f x)}\right )^3 \tan ^{-1}\left (e^{i (e+f x)}\right )\right ) (a+i a \tan (e+f x))^{3/2}}{3 f \sec ^{\frac{3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-I/3)*(c/(1 + E^((2*I)*(e + f*x))))^(5/2)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*(E^(I*(e + f*x))*(
-3 + 8*E^((2*I)*(e + f*x)) + 3*E^((4*I)*(e + f*x))) + 3*(1 + E^((2*I)*(e + f*x)))^3*ArcTan[E^(I*(e + f*x))])*(
a + I*a*Tan[e + f*x])^(3/2))/(E^((2*I)*(e + f*x))*f*Sec[e + f*x]^(3/2))

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Maple [A]  time = 0.03, size = 186, normalized size = 1.2 \begin{align*}{\frac{a{c}^{2}}{6\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( -2\,i \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+3\,ac\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) -2\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+3\,\tan \left ( fx+e \right ) \sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^2*a*(-2*I*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^
(1/2)*(a*c)^(1/2)+3*a*c*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))-2*I*(a*c*(1+
tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+3*tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))
^(1/2)/(a*c)^(1/2)

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Maxima [B]  time = 2.49565, size = 1202, normalized size = 7.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-(12*a*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*a*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) - 12*a*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a*c^2*sin(5/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*I*a*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*I
*a*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (6*a*c^2*cos(6*f*x + 6*e) + 18*a*c^2*cos(4*f*x +
4*e) + 18*a*c^2*cos(2*f*x + 2*e) + 6*I*a*c^2*sin(6*f*x + 6*e) + 18*I*a*c^2*sin(4*f*x + 4*e) + 18*I*a*c^2*sin(
2*f*x + 2*e) + 6*a*c^2)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e))) + 1) + (6*a*c^2*cos(6*f*x + 6*e) + 18*a*c^2*cos(4*f*x + 4*e) + 18*a*c^2*cos(2*f*x
+ 2*e) + 6*I*a*c^2*sin(6*f*x + 6*e) + 18*I*a*c^2*sin(4*f*x + 4*e) + 18*I*a*c^2*sin(2*f*x + 2*e) + 6*a*c^2)*ar
ctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))) + 1) - (-3*I*a*c^2*cos(6*f*x + 6*e) - 9*I*a*c^2*cos(4*f*x + 4*e) - 9*I*a*c^2*cos(2*f*x + 2*e) + 3*a*c^2*si
n(6*f*x + 6*e) + 9*a*c^2*sin(4*f*x + 4*e) + 9*a*c^2*sin(2*f*x + 2*e) - 3*I*a*c^2)*log(cos(1/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (3*I*a*c^2*cos(6*f*x + 6*e) + 9*I*a*c^2*cos(4*f*x + 4*e) + 9*I*a*c^2
*cos(2*f*x + 2*e) - 3*a*c^2*sin(6*f*x + 6*e) - 9*a*c^2*sin(4*f*x + 4*e) - 9*a*c^2*sin(2*f*x + 2*e) + 3*I*a*c^2
)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/(f*(-12*I*cos(6*f*x + 6
*e) - 36*I*cos(4*f*x + 4*e) - 36*I*cos(2*f*x + 2*e) + 12*sin(6*f*x + 6*e) + 36*sin(4*f*x + 4*e) + 36*sin(2*f*x
+ 2*e) - 12*I))

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Fricas [B]  time = 1.57819, size = 1108, normalized size = 6.97 \begin{align*} \frac{2 \,{\left (-6 i \, a c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 16 i \, a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} - 3 \, \sqrt{\frac{a^{3} c^{5}}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + \sqrt{\frac{a^{3} c^{5}}{f^{2}}}{\left (4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, f\right )}}{a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2}}\right ) + 3 \, \sqrt{\frac{a^{3} c^{5}}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + \sqrt{\frac{a^{3} c^{5}}{f^{2}}}{\left (-4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, f\right )}}{a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2}}\right )}{12 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/12*(2*(-6*I*a*c^2*e^(4*I*f*x + 4*I*e) - 16*I*a*c^2*e^(2*I*f*x + 2*I*e) + 6*I*a*c^2)*sqrt(a/(e^(2*I*f*x + 2*I
*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - 3*sqrt(a^3*c^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f
*e^(2*I*f*x + 2*I*e) + f)*log((8*(a*c^2*e^(2*I*f*x + 2*I*e) + a*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/
(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + sqrt(a^3*c^5/f^2)*(4*I*f*e^(2*I*f*x + 2*I*e) - 4*I*f))/(a*c^2*e^(
2*I*f*x + 2*I*e) + a*c^2)) + 3*sqrt(a^3*c^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((8*
(a*c^2*e^(2*I*f*x + 2*I*e) + a*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f
*x + I*e) + sqrt(a^3*c^5/f^2)*(-4*I*f*e^(2*I*f*x + 2*I*e) + 4*I*f))/(a*c^2*e^(2*I*f*x + 2*I*e) + a*c^2)))/(f*e
^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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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+I*a*tan(f*x+e))**(3/2)*(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{3}{2}}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(3/2)*(-I*c*tan(f*x + e) + c)^(5/2), x)