3.11.42 \(\int \frac {1}{(a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) [1042]

Optimal. Leaf size=200 \[ \frac {i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}}+\frac {6 \tan (e+f x)}{35 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {8 \tan (e+f x)}{35 a^2 c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {16 \tan (e+f x)}{35 a^3 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

16/35*tan(f*x+e)/a^3/c^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1/7*I/f/(a+I*a*tan(f*x+e))^(7/2)/
(c-I*c*tan(f*x+e))^(5/2)+6/35*tan(f*x+e)/a/f/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2)+8/35*tan(f*x+e)
/a^2/c/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.12, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3604, 47, 40, 39} \begin {gather*} \frac {16 \tan (e+f x)}{35 a^3 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {8 \tan (e+f x)}{35 a^2 c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {6 \tan (e+f x)}{35 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(I/7)/(f*(a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(5/2)) + (6*Tan[e + f*x])/(35*a*f*(a + I*a*Tan[e
+ f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)) + (8*Tan[e + f*x])/(35*a^2*c*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I
*c*Tan[e + f*x])^(3/2)) + (16*Tan[e + f*x])/(35*a^3*c^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]
])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-x)*(a + b*x)^(m + 1)*((c + d*x)^(m
+ 1)/(2*a*c*(m + 1))), x] + Dist[(2*m + 3)/(2*a*c*(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] && ILtQ[m + 3/2, 0]

Rule 47

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

Rule 3604

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{9/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}}+\frac {(6 c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}}+\frac {6 \tan (e+f x)}{35 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {24 \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{35 a f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}}+\frac {6 \tan (e+f x)}{35 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {8 \tan (e+f x)}{35 a^2 c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {16 \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{35 a^2 c f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{5/2}}+\frac {6 \tan (e+f x)}{35 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {8 \tan (e+f x)}{35 a^2 c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {16 \tan (e+f x)}{35 a^3 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 5.53, size = 115, normalized size = 0.58 \begin {gather*} -\frac {i (-350+175 \cos (2 (e+f x))+14 \cos (4 (e+f x))+\cos (6 (e+f x))+350 i \sin (2 (e+f x))+56 i \sin (4 (e+f x))+6 i \sin (6 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{1120 a^3 c^3 f \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

((-1/1120*I)*(-350 + 175*Cos[2*(e + f*x)] + 14*Cos[4*(e + f*x)] + Cos[6*(e + f*x)] + (350*I)*Sin[2*(e + f*x)]
+ (56*I)*Sin[4*(e + f*x)] + (6*I)*Sin[6*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*c^3*f*Sqrt[a + I*a*Tan[e
+ f*x]])

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Maple [A]
time = 0.37, size = 151, normalized size = 0.76

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (16 i \left (\tan ^{7}\left (f x +e \right )\right )-16 \left (\tan ^{8}\left (f x +e \right )\right )+56 i \left (\tan ^{5}\left (f x +e \right )\right )-56 \left (\tan ^{6}\left (f x +e \right )\right )+70 i \left (\tan ^{3}\left (f x +e \right )\right )-70 \left (\tan ^{4}\left (f x +e \right )\right )+30 i \tan \left (f x +e \right )-35 \left (\tan ^{2}\left (f x +e \right )\right )-5\right )}{35 f \,a^{4} c^{3} \left (-\tan \left (f x +e \right )+i\right )^{5} \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(151\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (16 i \left (\tan ^{7}\left (f x +e \right )\right )-16 \left (\tan ^{8}\left (f x +e \right )\right )+56 i \left (\tan ^{5}\left (f x +e \right )\right )-56 \left (\tan ^{6}\left (f x +e \right )\right )+70 i \left (\tan ^{3}\left (f x +e \right )\right )-70 \left (\tan ^{4}\left (f x +e \right )\right )+30 i \tan \left (f x +e \right )-35 \left (\tan ^{2}\left (f x +e \right )\right )-5\right )}{35 f \,a^{4} c^{3} \left (-\tan \left (f x +e \right )+i\right )^{5} \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/35/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^4/c^3*(16*I*tan(f*x+e)^7-16*tan(f*x+e)^8+56*I*
tan(f*x+e)^5-56*tan(f*x+e)^6+70*I*tan(f*x+e)^3-70*tan(f*x+e)^4+30*I*tan(f*x+e)-35*tan(f*x+e)^2-5)/(-tan(f*x+e)
+I)^5/(tan(f*x+e)+I)^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 0.89, size = 167, normalized size = 0.84 \begin {gather*} \frac {\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}} {\left (-7 i \, e^{\left (14 i \, f x + 14 i \, e\right )} - 77 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 595 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 320 i \, e^{\left (9 i \, f x + 9 i \, e\right )} + 175 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 320 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 875 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 217 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 47 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-7 i \, f x - 7 i \, e\right )}}{2240 \, a^{4} c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/2240*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-7*I*e^(14*I*f*x + 14*I*e) - 77*I*
e^(12*I*f*x + 12*I*e) - 595*I*e^(10*I*f*x + 10*I*e) - 320*I*e^(9*I*f*x + 9*I*e) + 175*I*e^(8*I*f*x + 8*I*e) -
320*I*e^(7*I*f*x + 7*I*e) + 875*I*e^(6*I*f*x + 6*I*e) + 217*I*e^(4*I*f*x + 4*I*e) + 47*I*e^(2*I*f*x + 2*I*e) +
5*I)*e^(-7*I*f*x - 7*I*e)/(a^4*c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**(7/2)/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4848 deep

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Giac [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(1/(a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(7/2)*(-I*c*tan(f*x + e) + c)^(5/2)), x)

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Mupad [B]
time = 6.70, size = 186, normalized size = 0.93 \begin {gather*} \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,630{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,168{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,42{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,5{}\mathrm {i}+770\,\sin \left (2\,e+2\,f\,x\right )+182\,\sin \left (4\,e+4\,f\,x\right )+42\,\sin \left (6\,e+6\,f\,x\right )+5\,\sin \left (8\,e+8\,f\,x\right )-525{}\mathrm {i}\right )}{2240\,a^4\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)^(7/2)*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*630i + cos(
4*e + 4*f*x)*168i + cos(6*e + 6*f*x)*42i + cos(8*e + 8*f*x)*5i + 770*sin(2*e + 2*f*x) + 182*sin(4*e + 4*f*x) +
42*sin(6*e + 6*f*x) + 5*sin(8*e + 8*f*x) - 525i))/(2240*a^4*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i
+ 1))/(cos(2*e + 2*f*x) + 1))^(1/2))

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