3.1.87 \(\int \frac {\csc ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) [87]

Optimal. Leaf size=43 \[ -\frac {2 d^3}{7 b (d \tan (a+b x))^{7/2}}-\frac {2 d}{3 b (d \tan (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2671, 14} \begin {gather*} -\frac {2 d^3}{7 b (d \tan (a+b x))^{7/2}}-\frac {2 d}{3 b (d \tan (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^4/Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*d^3)/(7*b*(d*Tan[a + b*x])^(7/2)) - (2*d)/(3*b*(d*Tan[a + b*x])^(3/2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx &=\frac {d \text {Subst}\left (\int \frac {d^2+x^2}{x^{9/2}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac {d \text {Subst}\left (\int \left (\frac {d^2}{x^{9/2}}+\frac {1}{x^{5/2}}\right ) \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {2 d^3}{7 b (d \tan (a+b x))^{7/2}}-\frac {2 d}{3 b (d \tan (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 40, normalized size = 0.93 \begin {gather*} \frac {2 d (-5+2 \cos (2 (a+b x))) \csc ^2(a+b x)}{21 b (d \tan (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^4/Sqrt[d*Tan[a + b*x]],x]

[Out]

(2*d*(-5 + 2*Cos[2*(a + b*x)])*Csc[a + b*x]^2)/(21*b*(d*Tan[a + b*x])^(3/2))

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Maple [A]
time = 0.61, size = 50, normalized size = 1.16

method	result	size
default	\(\frac {2 \left (4 \left (\cos ^{2}\left (b x +a \right )\right )-7\right ) \cos \left (b x +a \right )}{21 b \sin \left (b x +a \right )^{3} \sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^4/(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/21/b*(4*cos(b*x+a)^2-7)*cos(b*x+a)/sin(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(1/2)

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Maxima [A]
time = 0.27, size = 35, normalized size = 0.81 \begin {gather*} -\frac {2 \, {\left (7 \, d^{2} \tan \left (b x + a\right )^{2} + 3 \, d^{2}\right )} d}{21 \, \left (d \tan \left (b x + a\right )\right )^{\frac {7}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^4/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

-2/21*(7*d^2*tan(b*x + a)^2 + 3*d^2)*d/((d*tan(b*x + a))^(7/2)*b)

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Fricas [A]
time = 0.41, size = 70, normalized size = 1.63 \begin {gather*} \frac {2 \, {\left (4 \, \cos \left (b x + a\right )^{4} - 7 \, \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{21 \, {\left (b d \cos \left (b x + a\right )^{4} - 2 \, b d \cos \left (b x + a\right )^{2} + b d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^4/(d*tan(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

2/21*(4*cos(b*x + a)^4 - 7*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*d*cos(b*x + a)^4 - 2*b*d*cos(b
*x + a)^2 + b*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{4}{\left (a + b x \right )}}{\sqrt {d \tan {\left (a + b x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**4/(d*tan(b*x+a))**(1/2),x)

[Out]

Integral(csc(a + b*x)**4/sqrt(d*tan(a + b*x)), x)

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Giac [A]
time = 0.58, size = 45, normalized size = 1.05 \begin {gather*} -\frac {2 \, {\left (7 \, d^{3} \tan \left (b x + a\right )^{2} + 3 \, d^{3}\right )}}{21 \, \sqrt {d \tan \left (b x + a\right )} b d^{3} \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)^4/(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

-2/21*(7*d^3*tan(b*x + a)^2 + 3*d^3)/(sqrt(d*tan(b*x + a))*b*d^3*tan(b*x + a)^3)

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Mupad [B]
time = 7.19, size = 530, normalized size = 12.33 \begin {gather*} \frac {344\,\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}}}{105\,b\,d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}+\frac {40\,\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}}}{21\,b\,d\,{\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\,{\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}}\,304{}\mathrm {i}}{105\,b\,d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}+\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\,{\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\,{\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\,{\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)^4*(d*tan(a + b*x))^(1/2)),x)

[Out]

(344*(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))/(105*b*d*(ex
p(a*2i + b*x*2i) - 1)) + (40*(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))/(21*b*d*(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*(exp(a*2i + b*x*2i) - 1)^3) - ((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)*304i)/(105*b*d*(exp(a*2i + b*x*2i)*1i - 1i)) + (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*(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*(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*(exp(a*2i + b*x*2i)*1i - 1i)^4)

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