Optimal. Leaf size=216 \[ -\frac {32 \sqrt {\pi } b^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}-\frac {32 \sqrt {\pi } b^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}} \]
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Rubi [A] time = 0.34, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3314, 32, 3313, 12, 3306, 3305, 3351, 3304, 3352} \[ -\frac {32 \sqrt {\pi } b^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{15 d^{7/2}}-\frac {32 \sqrt {\pi } b^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 32
Rule 3304
Rule 3305
Rule 3306
Rule 3313
Rule 3314
Rule 3351
Rule 3352
Rubi steps
\begin {align*} \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx &=-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {\left (8 b^2\right ) \int \frac {1}{(c+d x)^{3/2}} \, dx}{15 d^2}-\frac {\left (16 b^2\right ) \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {\left (64 b^3\right ) \int \frac {\sin (2 a+2 b x)}{2 \sqrt {c+d x}} \, dx}{15 d^3}\\ &=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {\left (32 b^3\right ) \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{15 d^3}\\ &=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {\left (32 b^3 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx}{15 d^3}-\frac {\left (32 b^3 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx}{15 d^3}\\ &=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {\left (64 b^3 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{15 d^4}-\frac {\left (64 b^3 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{15 d^4}\\ &=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {32 b^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}-\frac {32 b^{5/2} \sqrt {\pi } C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{15 d^{7/2}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}\\ \end {align*}
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Mathematica [A] time = 2.13, size = 244, normalized size = 1.13 \[ -\frac {16 b^2 c^2 \cos (2 (a+b x))+32 b^2 c d x \cos (2 (a+b x))+16 b^2 d^2 x^2 \cos (2 (a+b x))+32 \sqrt {\pi } b d \left (\frac {b}{d}\right )^{3/2} (c+d x)^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+32 \sqrt {\pi } b d \left (\frac {b}{d}\right )^{3/2} (c+d x)^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+4 b c d \sin (2 (a+b x))+4 b d^2 x \sin (2 (a+b x))-3 d^2 \cos (2 (a+b x))+3 d^2}{15 d^3 (c+d x)^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 328, normalized size = 1.52 \[ -\frac {2 \, {\left (16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - {\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 3 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 \, d^{2}\right )} \sqrt {d x + c}\right )}}{15 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 230, normalized size = 1.06 \[ \frac {-\frac {1}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {4 b \left (-\frac {\sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.97, size = 135, normalized size = 0.62 \[ \frac {\sqrt {2} {\left ({\left (-\left (5 i + 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (5 i - 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (5 i - 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (5 i + 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} - 2}{10 \, {\left (d x + c\right )}^{\frac {5}{2}} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{7/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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