3.1.74 \(\int (c+d x)^m \sinh ^2(a+b x) \, dx\) [74]

Optimal. Leaf size=144 \[ -\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 a+\frac {2 b c}{d}} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b} \]

[Out]

-1/2*(d*x+c)^(1+m)/d/(1+m)+2^(-3-m)*exp(2*a-2*b*c/d)*(d*x+c)^m*GAMMA(1+m,-2*b*(d*x+c)/d)/b/((-b*(d*x+c)/d)^m)-
2^(-3-m)*exp(-2*a+2*b*c/d)*(d*x+c)^m*GAMMA(1+m,2*b*(d*x+c)/d)/b/((b*(d*x+c)/d)^m)

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Rubi [A]
time = 0.14, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3393, 3388, 2212} \begin {gather*} \frac {2^{-m-3} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m-3} e^{\frac {2 b c}{d}-2 a} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 b (c+d x)}{d}\right )}{b}-\frac {(c+d x)^{m+1}}{2 d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*Sinh[a + b*x]^2,x]

[Out]

-1/2*(c + d*x)^(1 + m)/(d*(1 + m)) + (2^(-3 - m)*E^(2*a - (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (-2*b*(c + d*x))
/d])/(b*(-((b*(c + d*x))/d))^m) - (2^(-3 - m)*E^(-2*a + (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (2*b*(c + d*x))/d]
)/(b*((b*(c + d*x))/d)^m)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin {align*} \int (c+d x)^m \sinh ^2(a+b x) \, dx &=-\int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cosh (2 a+2 b x)\right ) \, dx\\ &=-\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {1}{2} \int (c+d x)^m \cosh (2 a+2 b x) \, dx\\ &=-\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} (c+d x)^m \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} (c+d x)^m \, dx\\ &=-\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 a+\frac {2 b c}{d}} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 131, normalized size = 0.91 \begin {gather*} \frac {1}{8} (c+d x)^m \left (-\frac {4 (c+d x)}{d (1+m)}+\frac {2^{-m} e^{2 a-\frac {2 b c}{d}} \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m} e^{-2 a+\frac {2 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m*Sinh[a + b*x]^2,x]

[Out]

((c + d*x)^m*((-4*(c + d*x))/(d*(1 + m)) + (E^(2*a - (2*b*c)/d)*Gamma[1 + m, (-2*b*(c + d*x))/d])/(2^m*b*(-((b
*(c + d*x))/d))^m) - (E^(-2*a + (2*b*c)/d)*Gamma[1 + m, (2*b*(c + d*x))/d])/(2^m*b*((b*(c + d*x))/d)^m)))/8

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{m} \left (\sinh ^{2}\left (b x +a \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m*sinh(b*x+a)^2,x)

[Out]

int((d*x+c)^m*sinh(b*x+a)^2,x)

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Maxima [A]
time = 0.07, size = 102, normalized size = 0.71 \begin {gather*} -\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1}}{2 \, d {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/4*(d*x + c)^(m + 1)*e^(-2*a + 2*b*c/d)*exp_integral_e(-m, 2*(d*x + c)*b/d)/d - 1/4*(d*x + c)^(m + 1)*e^(2*a
- 2*b*c/d)*exp_integral_e(-m, -2*(d*x + c)*b/d)/d - 1/2*(d*x + c)^(m + 1)/(d*(m + 1))

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Fricas [A]
time = 0.09, size = 241, normalized size = 1.67 \begin {gather*} -\frac {{\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) + {\left (d m + d\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) + 4 \, {\left (b d x + b c\right )} \cosh \left (m \log \left (d x + c\right )\right ) + 4 \, {\left (b d x + b c\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \, {\left (b d m + b d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/8*((d*m + d)*cosh((d*m*log(2*b/d) - 2*b*c + 2*a*d)/d)*gamma(m + 1, 2*(b*d*x + b*c)/d) - (d*m + d)*cosh((d*m
*log(-2*b/d) + 2*b*c - 2*a*d)/d)*gamma(m + 1, -2*(b*d*x + b*c)/d) - (d*m + d)*gamma(m + 1, 2*(b*d*x + b*c)/d)*
sinh((d*m*log(2*b/d) - 2*b*c + 2*a*d)/d) + (d*m + d)*gamma(m + 1, -2*(b*d*x + b*c)/d)*sinh((d*m*log(-2*b/d) +
2*b*c - 2*a*d)/d) + 4*(b*d*x + b*c)*cosh(m*log(d*x + c)) + 4*(b*d*x + b*c)*sinh(m*log(d*x + c)))/(b*d*m + b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m*sinh(b*x+a)**2,x)

[Out]

Integral((c + d*x)**m*sinh(a + b*x)**2, x)

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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((d*x+c)^m*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^m*sinh(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^2*(c + d*x)^m,x)

[Out]

int(sinh(a + b*x)^2*(c + d*x)^m, x)

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