3.2.5 \(\int \frac {\csc ^2(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) [105]

Optimal. Leaf size=20 \[ -\frac {2 d}{7 b (d \tan (a+b x))^{7/2}} \]

[Out]

-2/7*d/b/(d*tan(b*x+a))^(7/2)

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Rubi [A]
time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2671, 30} \begin {gather*} -\frac {2 d}{7 b (d \tan (a+b x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2/(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*d)/(7*b*(d*Tan[a + b*x])^(7/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\csc ^2(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=\frac {d \text {Subst}\left (\int \frac {1}{x^{9/2}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {2 d}{7 b (d \tan (a+b x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 20, normalized size = 1.00 \begin {gather*} -\frac {2 d}{7 b (d \tan (a+b x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2/(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*d)/(7*b*(d*Tan[a + b*x])^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(37\) vs. \(2(16)=32\).
time = 0.34, size = 38, normalized size = 1.90

method	result	size
default	\(-\frac {2 \cos \left (b x +a \right )}{7 b \sin \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2/(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/7/b*cos(b*x+a)/sin(b*x+a)/(d*sin(b*x+a)/cos(b*x+a))^(5/2)

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Maxima [A]
time = 0.27, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2}{7 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} b \tan \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/7/((d*tan(b*x + a))^(5/2)*b*tan(b*x + a))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (16) = 32\).
time = 0.39, size = 63, normalized size = 3.15 \begin {gather*} -\frac {2 \, \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )^{4}}{7 \, {\left (b d^{3} \cos \left (b x + a\right )^{4} - 2 \, b d^{3} \cos \left (b x + a\right )^{2} + b d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

-2/7*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b*x + a)^4/(b*d^3*cos(b*x + a)^4 - 2*b*d^3*cos(b*x + a)^2 + b*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{2}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2/(d*tan(b*x+a))**(5/2),x)

[Out]

Integral(csc(a + b*x)**2/(d*tan(a + b*x))**(5/2), x)

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Giac [A]
time = 0.62, size = 26, normalized size = 1.30 \begin {gather*} -\frac {2}{7 \, \sqrt {d \tan \left (b x + a\right )} b d^{2} \tan \left (b x + a\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2/(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/7/(sqrt(d*tan(b*x + a))*b*d^2*tan(b*x + a)^3)

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Mupad [B]
time = 7.40, size = 530, normalized size = 26.50 \begin {gather*} \frac {46\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{7\,b\,d^3\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}+\frac {12\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{5\,b\,d^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^2}+\frac {24\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{35\,b\,d^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3}-\frac {\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}\,48{}\mathrm {i}}{7\,b\,d^3\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}+\frac {144\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{35\,b\,d^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}\,144{}\mathrm {i}}{35\,b\,d^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {16\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{7\,b\,d^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^2*(d*tan(a + b*x))^(5/2)),x)

[Out]

(46*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(7*b*d^3*(exp
(a*2i + b*x*2i) - 1)) + (12*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) +
1))^(1/2))/(5*b*d^3*(exp(a*2i + b*x*2i) - 1)^2) + (24*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1
i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(35*b*d^3*(exp(a*2i + b*x*2i) - 1)^3) - ((exp(a*2i + b*x*2i) + 1)*(-(d*(e
xp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2)*48i)/(7*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)) + (144
*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(35*b*d^3*(exp(a
*2i + b*x*2i)*1i - 1i)^2) + ((exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) +
1))^(1/2)*144i)/(35*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^3) - (16*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x
*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(7*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^4)

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